CBSE NCERT Solutions for Class 12 Maths Chapter 07 · CBSE NCERT Solutions for Class 12 Maths...

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CBSE NCERT Solutions for Class 12 Maths Chapter 07

Back of Chapter Questions

Exercise 7.1

1. Find an anti-derivative (or integral) of the function sin 2𝑥 by the method of inspection.

[2 Marks]

Solution:

Anti-derivative of sin 2𝑥 is a function of 𝑥 whose derivative is sin 2𝑥.

As we know that, 𝑑

𝑑𝑥(cos 2𝑥) = −2 sin 2𝑥 [1 Mark]

sin 2𝑥 = −1

2

𝑑

𝑑𝑥(cos 2𝑥)

∴ sin 2𝑥 =𝑑

𝑑𝑥(−

1

2cos 2𝑥)

Hence, the anti – derivative of sin 2𝑥 is −1

2cos 2𝑥 [1 Mark]

2. Find an anti-derivative (or integral) of the function cos 3𝑥 by the method of inspection.

[2 Marks]

Solution:

Anti-derivative of cos 3𝑥 is a function of 𝑥 whose derivative is cos 3𝑥.


𝑑𝑥(sin 3𝑥) = 3 cos 3𝑥 [1 Mark]

⇒ cos 3𝑥 =1

3

𝑑

𝑑𝑥(sin 3𝑥)

∴ cos 3𝑥 =𝑑

𝑑𝑥(

1

3sin 3𝑥)

Hence, the anti-derivative of cos 3𝑥 is 1

3sin 3𝑥 [1 Mark]



3. Find an anti-derivative (or integral) of the function 𝑒2𝑥 by the method of inspection.

[2 Marks]

Solution:

Anti-derivative of 𝑒2𝑥 is the function of 𝑥 whose derivative is 𝑒2𝑥.


𝑑𝑥(𝑒2𝑥) = 2𝑒2𝑥 [1 Mark]

⇒ 𝑒2𝑥 =1

2

𝑑

𝑑𝑥(𝑒2𝑥)

∴ 𝑒2𝑥 =1

2

𝑑

𝑑𝑥(𝑒2𝑥)

Hence, the anti-derivative of 𝑒2𝑥 is 1

2𝑒2𝑥 [1 Mark]

4. Find an anti-derivative (or integral) of the function (𝑎𝑥 + 𝑏)2 by the method of inspection.

[2 Marks]

Solution:

Anti-derivative of (𝑎𝑥 + 𝑏)2 is the function of 𝑥 whose derivative is (𝑎𝑥 + 𝑏)2.


𝑑𝑥(𝑎𝑥 + 𝑏)3 = 3𝑎(𝑎𝑥 + 𝑏)2 [1 Mark]

⇒ (𝑎𝑥 + 𝑏)2 =1

3𝑎

𝑑

𝑑𝑥(𝑎𝑥 + 𝑏)3

∴ (𝑎𝑥 + 𝑏)2 =𝑑

𝑑𝑥(

1

3𝑎(𝑎𝑥 + 𝑏)3)

Hence, the anti-derivative of (𝑎𝑥 + 𝑏)2 is 1

3𝑎(𝑎𝑥 + 𝑏)3 [1 Mark]

5. Find an anti-derivative (or integral) of the function sin 2𝑥 – 4𝑒3𝑥 by the method of inspection.

[2 Marks]



Solution:

Anti-derivative of sin 2𝑥 – 4𝑒3𝑥 is the function of 𝑥 whose derivative is sin 2𝑥 – 4𝑒3𝑥


𝑑𝑥(−

1

2cos 2𝑥 −

4

3𝑒3𝑥) = sin 2𝑥 − 4𝑒3𝑥 [1 Mark]

Hence, the anti-derivative of (sin 2𝑥 − 4𝑒2𝑥) is (−1

2cos 2𝑥 −

4

3𝑒3𝑥) [1 Mark]

6. Find the integral ∫(4𝑒3𝑥 + 1)𝑑𝑥. [2 Marks]

Solution:

The given integral is ∫(4𝑒3𝑥 + 1)𝑑𝑥

= 4 ∫ 𝑒3𝑥𝑑𝑥 + ∫ 1𝑑𝑥

= 4 (𝑒3𝑥

3) + 𝑥 + 𝐶 [∵ ∫ 𝑒𝑎𝑥𝑑𝑥 =

𝑒𝑎𝑥

𝑎+ 𝐶 𝑎𝑛𝑑 ∫ 𝑎 𝑑𝑥 = 𝑎𝑥 + 𝐶] [1 Mark]

=4

3𝑒3𝑥 + 𝑥 + 𝐶

Hence, integral ∫(4𝑒3𝑥 + 1)𝑑𝑥 =4

3𝑒3𝑥 + 𝑥 + 𝐶. [1 Mark]

7. Find the integral ∫ 𝑥2 (1 −1

𝑥2) 𝑑𝑥 [2 Marks]

Solution:

The given integral is ∫ 𝑥2 (1 −1

𝑥2) 𝑑𝑥

= ∫(𝑥2 − 1)𝑑𝑥

= ∫ 𝑥2𝑑𝑥 − ∫ 1𝑑𝑥

[𝟏

𝟐 Mark]

=𝑥3

3− 𝑥 + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶 𝑎𝑛𝑑 ∫ 𝑎 𝑑𝑥 = 𝑎𝑥 + 𝐶 ] [1 Mark]

Hence, the integral ∫ 𝑥2 (1 −1

𝑥2) 𝑑𝑥 = 𝑥3

3− 𝑥 + 𝐶

[𝟏

𝟐 Mark]



8. Find the integral ∫(𝑎𝑥2 + 𝑏𝑥 + 𝑐)𝑑𝑥 [2 Marks]

Solution:

The given integral is ∫(𝑎𝑥2 + 𝑏𝑥 + 𝑐)𝑑𝑥

= 𝑎 ∫ 𝑥2𝑑𝑥 + 𝑏 ∫ 𝑥𝑑𝑥 + 𝑐 ∫ 1. 𝑑𝑥

= 𝑎 (𝑥3

3) + 𝑏 (

𝑥2

2) + 𝑐𝑥 + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶 𝑎𝑛𝑑 ∫ 𝑎 𝑑𝑥 = 𝑎𝑥 + 𝐶]

[1 Mark]

=𝑎𝑥3

3+

𝑏𝑥2

2+ 𝑐𝑥 + 𝐶

Hence, the integral ∫(𝑎𝑥2 + 𝑏𝑥 + 𝑐)𝑑𝑥 = 𝑎𝑥3

3+

𝑏𝑥2

2+ 𝑐𝑥 + 𝐶

[1 Mark]

9. Find the integral ∫(2𝑥2 + 𝑒𝑥) 𝑑𝑥. [2 Marks]

Solution:

The given integral is ∫(2𝑥2 + 𝑒𝑥) 𝑑𝑥

= 2 ∫ 𝑥2𝑑𝑥 + ∫ 𝑒𝑥𝑑𝑥

= 2 (𝑥3

3) + 𝑒𝑥 + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶 𝑎𝑛𝑑 ∫ 𝑒𝑎𝑥𝑑𝑥 =

𝑒𝑎𝑥

𝑎+ 𝐶]

[1 Mark]

=2

3𝑥3 + 𝑒𝑥 + 𝐶

Hence, the integral ∫(2𝑥2 + 𝑒𝑥) 𝑑𝑥 =2

3𝑥3 + 𝑒𝑥 + 𝐶.

[1 Mark]

10. Find the integral ∫ (√𝑥 −1

√𝑥)

2𝑑𝑥 [2 Marks]



Solution:

The given integral is ∫ (√𝑥 −1

√𝑥)

2𝑑𝑥

= ∫ (𝑥 +1

𝑥− 2)

𝑑𝑥

= ∫ 𝑥𝑑𝑥 + ∫1

𝑥𝑑𝑥 − 2 ∫ 1. 𝑑𝑥

[𝟏

𝟐 Mark]

=𝑥2

2+ log|𝑥| − 2𝑥 + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶, ∫ 𝑎 𝑑𝑥 = 𝑎𝑥 + 𝐶 𝑎𝑛𝑑 ∫

1

𝑥𝑑𝑥 = log|𝑥| + 𝐶 ]

[1 Mark]

Hence, the integral ∫ (√𝑥 −1

√𝑥)

2𝑑𝑥 =

𝑥2

2+ log|𝑥| − 2𝑥 + 𝐶

[𝟏

𝟐 Mark]

11. Find the integral ∫𝑥3+5𝑥2−4

𝑥2 𝑑𝑥 [2 Marks]

Solution:

The given integral is ∫𝑥3+5𝑥2−4

𝑥2 𝑑𝑥

= ∫(𝑥 + 5 − 4𝑥−2)𝑑𝑥

= ∫ 𝑥 𝑑𝑥 + 5 ∫ 1. 𝑑𝑥 − 4 ∫ 𝑥−2𝑑𝑥

[𝟏

𝟐 Mark]

=𝑥2

2+ 5𝑥 − 4 (

𝑥−1

−1) + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶]

[1 Mark]

=𝑥2

2+ 5𝑥 +

4

𝑥+ 𝐶

Hence, the integral ∫𝑥3+5𝑥2−4

𝑥2 𝑑𝑥 =𝑥2

2+ 5𝑥 +

4

𝑥+ 𝐶

[𝟏

𝟐 Mark]



12. Find the integral ∫𝑥3+3𝑥+4

√𝑥𝑑𝑥 [2 Marks]

Solution:

The given integral is ∫𝑥3+3𝑥+4

√𝑥𝑑𝑥

= ∫ (𝑥5

2 + 3𝑥1

2 + 4𝑥−1

2) 𝑑𝑥

=𝑥7

2

7

2 +3(𝑥

32)

3

2

+4(𝑥

12)

1

2

+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶 ] [1 Mark]

=2

7𝑥

7

2 + 2𝑥3

2 + 8𝑥1

2 + 𝐶

=2

7𝑥

7

2 + 2𝑥3

2 + 8√𝑥 + 𝐶

Hence, the integral ∫𝑥3+3𝑥+4

√𝑥𝑑𝑥 =

2

7𝑥

7

2 + 2𝑥3

2 + 8√𝑥 + 𝐶 [1 Mark]

13. Find the integral ∫𝑥3−𝑥2+𝑥−1

𝑥−1𝑑𝑥 [2 Marks]

Solution:

The given integral is ∫𝑥3−𝑥2+𝑥−1

𝑥−1𝑑𝑥

On dividing, we get

= ∫(𝑥2 + 1)𝑑𝑥 [𝟏

𝟐 Mark]

= ∫ 𝑥2𝑑𝑥 + ∫ 1𝑑𝑥

=𝑥3

3+ 𝑥 + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶 𝑎𝑛𝑑 ∫ 𝑎 𝑑𝑥 = 𝑎𝑥 + 𝐶] [1 Mark]

Hence, the integral ∫𝑥3−𝑥2+𝑥−1

𝑥−1𝑑𝑥 =

𝑥3

3+ 𝑥 + 𝐶 [

𝟏

𝟐 Mark]

14. Find the integral ∫(1 − 𝑥)√𝑥𝑑𝑥 [2 Marks]



Solution:

The given integral is ∫(1 − 𝑥)√𝑥𝑑𝑥

= ∫ (√𝑥 − 𝑥3

2) 𝑑𝑥

[𝟏

𝟐 Mark]

=𝑥

32

3

2

−𝑥

52

5

2

+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

=2

3𝑥

3

2 −2

5𝑥

5

2 + 𝐶 [1 Mark]

Hence, the integral ∫(1 − 𝑥)√𝑥𝑑𝑥 =2

3𝑥

3

2 −2

5𝑥

5

2 + 𝐶

[𝟏

𝟐 Mark]

15. Find the integral ∫ √𝑥(3𝑥2 + 2𝑥 + 3)𝑑𝑥 [2 Marks]

Solution:

The given integral is ∫ √𝑥(3𝑥2 + 2𝑥 + 3)𝑑𝑥

= ∫ (3𝑥5

2 + 2𝑥3

2 + 3𝑥1

2) 𝑑𝑥

[𝟏

𝟐 Mark]

= 3 (𝑥

72

7

2

) + 2 (𝑥

52

5

2

) +3(𝑥

32)

3

2

+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

=6

7𝑥

7

2 +4

5𝑥

5

2 + 2𝑥3

2 + 𝐶

Hence, the integral ∫ √𝑥(3𝑥2 + 2𝑥 + 3)𝑑𝑥 =6

7𝑥

7

2 +4

5𝑥

5

2 + 2𝑥3

2 + 𝐶

[𝟏

𝟐 Mark]

16. Find the integral ∫(2𝑥 − 3 cos 𝑥 + 𝑒𝑥)𝑑𝑥 [4 Marks]

Solution:



The given integral is ∫(2𝑥 − 3 cos 𝑥 + 𝑒𝑥)𝑑𝑥

= 2 ∫ 𝑥𝑑𝑥 − 3 ∫ cos 𝑥 𝑑𝑥 + ∫ 𝑒𝑥𝑑𝑥

[𝟏

𝟐 Mark]

=2𝑥2

2− 3(sin 𝑥) + 𝑒𝑥 + 𝐶 [3 Marks]

[∵ ∫ 𝑥𝑛𝑥𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶 , ∫ 𝑒𝑎𝑥𝑑𝑥 =

𝑒𝑎𝑥

𝑎+ 𝐶 and ∫ 𝑐𝑜𝑠𝑥 𝑑𝑥 = 𝑠𝑖𝑛𝑥 + 𝐶]

= 𝑥2 − 3 sin 𝑥 + 𝑒𝑥 + 𝐶

Hence, the integral ∫(2𝑥 − 3 cos 𝑥 + 𝑒𝑥)𝑑𝑥 = 𝑥2 − 3 sin 𝑥 + 𝑒𝑥 + 𝐶

[𝟏

𝟐 Mark]

17. Find the integral ∫(2𝑥2 − 3 sin 𝑥 + 5√𝑥)𝑑𝑥 [4 Marks]

Solution:

The given integral is ∫(2𝑥2 − 3 sin 𝑥 + 5√𝑥)𝑑𝑥

= 2 ∫ 𝑥2𝑑𝑥 − 3 ∫ sin 𝑥𝑑𝑥 + 5 ∫ 𝑥1

2𝑑𝑥

[𝟏

𝟐 Mark]

= 2𝑥3

3− 3(− cos 𝑥) + 5 (

𝑥32

3

2

) + 𝐶

[∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶, and ∫ sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝐶] [3 Marks]

=2

3𝑥3 + 3 cos 𝑥 +

10

3𝑥

3

2 + 𝐶

Hence, the integral ∫(2𝑥2 − 3 sin 𝑥 + 5√𝑥)𝑑𝑥 =2

3𝑥3 + 3 cos 𝑥 +

10

3𝑥

3

2 + 𝐶

[𝟏

𝟐 Mark]

18. Find the integral ∫ sec 𝑥(sec 𝑥 + tan 𝑥)𝑑𝑥 [2 Marks]



Solution:

The given integral is ∫ sec 𝑥(sec 𝑥 + tan 𝑥)𝑑𝑥

= ∫(sec2 𝑥 + sec 𝑥 tan 𝑥)𝑑𝑥

[𝟏

𝟐 Mark]

= ∫ sec2 𝑥𝑑𝑥 + ∫ sec 𝑥 tan 𝑥𝑑𝑥

[∵ ∫ sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝐶 𝑎𝑛𝑑 ∫ sec2𝑥 𝑑𝑥 = tan 𝑥 + 𝐶]

[1 Mark]

= tan 𝑥 + sec 𝑥 + 𝐶

Hence, the integral ∫ sec 𝑥(sec 𝑥 + tan 𝑥)𝑑𝑥 = tan 𝑥 + sec 𝑥 + 𝐶

[𝟏

𝟐 Mark]

19. Find the integral ∫sec2 𝑥

cosec2𝑥𝑑𝑥 [3 Marks]

Solution:

The given integral is ∫sec2 𝑥

cosec2𝑥𝑑𝑥

= ∫

1

cos2 𝑥1

sin2 𝑥

𝑑𝑥

= ∫sin2 𝑥

cos2 𝑥𝑑𝑥

= ∫ tan2 𝑥 𝑑𝑥 [∵ ∫sin 𝑥

cos 𝑥= tan 𝑥 ]

[1 Mark]

= ∫(sec2 𝑥 − 1)𝑑𝑥 [∵ tan2𝑥 = sec2 𝑥 − 1 ]

[𝟏

𝟐 Mark]

= ∫ sec2 𝑥𝑑𝑥 − ∫ 1𝑑𝑥

= tan 𝑥 − 𝑥 + 𝐶 [∵ ∫ sec2𝑥𝑑𝑥 = tan 𝑥 + 𝐶]

[1 Marks]

Hence, the integral ∫sec2 𝑥

cosec2𝑥𝑑𝑥 = tan 𝑥 − 𝑥 + 𝐶



[𝟏

𝟐 Mark]

20. Find the integral ∫2−3 sin 𝑥

cos2 𝑥𝑑𝑥 [4 Marks]

Solution:

The given integral is ∫2−3 sin 𝑥

cos2 𝑥𝑑𝑥

= ∫ (2

cos2 𝑥−

3 sin 𝑥

cos2 𝑥) 𝑑𝑥

∫ 2 sec2 𝑥𝑑𝑥 − 3 ∫ tan 𝑥 sec 𝑥𝑑𝑥 [∵ sin 𝑥

cos 𝑥= tan 𝑥 𝑎𝑛𝑑

1

cos 𝑥= sec 𝑥]

[1 Mark]

= 2 tan 𝑥 − 3 sec 𝑥 + 𝐶

[2 Marks]

[∵ ∫ sec2𝑥 𝑑𝑥 = tan 𝑥 + 𝐶 𝑎𝑛𝑑 ∫ tan 𝑥 sec 𝑥𝑑𝑥 = sec 𝑥 + 𝐶]

Hence, the integral ∫2−3 sin 𝑥

cos2 𝑥𝑑𝑥 = 2 tan 𝑥 − 3 sec 𝑥 + 𝐶

[1 Mark]

21. The anti-derivative of (√𝑥 +1

√𝑥) equals

(A) 1

3𝑥

1

3 + 2𝑥1

2 + 𝐶

(B) 2

3𝑥

2

3 +1

2𝑥2 + 𝐶

(C) 2

3𝑥

3

2 + 2𝑥1

2 + 𝐶

(D) 3

2𝑥

3

2 +1

2𝑥

1

2 + 𝐶 [2 Marks]

Solution:

(√𝑥 +1

√𝑥) 𝑑𝑥



= ∫ 𝑥12 𝑑𝑥 + ∫ 𝑥−

12 𝑑𝑥

=𝑥

32

3

2

+𝑥

12

1

2

+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶]

=2

3𝑥

32 + 2𝑥

12 + 𝐶

The anti-derivative of (√𝑥 +1

√𝑥) =

2

3𝑥

3

2 + 2𝑥1

2 + 𝐶 [2 Marks]

Hence, the correct answer is (C)

22. If 𝑑

𝑑𝑥𝑓(𝑥) = 4𝑥3 −

3

𝑥4 such that 𝑓(2) = 0. Then 𝑓(𝑥) is

(A) 𝑥4 +1

𝑥3 −129

8

(B) 𝑥3 +1

𝑥4 +129

8

(C) 𝑥4 +1

𝑥3 +129

8

(D) 𝑥3 +1

𝑥4 −129

8 [4 Marks]

Solution:

Given that, 𝑑

𝑑𝑥𝑓(𝑥) = 4𝑥3 −

3

𝑥4

Thus, anti-derivative of 4𝑥3 −3

𝑥4 = 𝑓(𝑥) [1 Mark]

∴ 𝑓(𝑥) = ∫ 4𝑥3 −3

𝑥4𝑑𝑥

⇒ 𝑓(𝑥) = 4 ∫ 𝑥3𝑑𝑥 − 3 ∫(𝑥−4)𝑑𝑥

⇒ 𝑓(𝑥) = 4 (𝑥4

4) − 3 (

𝑥−3

−3) + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

⇒ 𝑓(𝑥) = 𝑥4 +1

𝑥3 + 𝐶

Also,

𝑓(2) = 0

∴ 𝑓(2) = (2)4 +1

(2)3 + 𝐶 = 0



⇒ 16 +1

8+ 𝐶 = 0

⇒ 𝐶 = − (16 +1

8)

⇒ 𝐶 =−129

8

[1 Mark]

∴ 𝑓(𝑥) = 𝑥4 +1

𝑥3 −129

8 [1 Mark]

Hence, the correct answer is (A)

Exercise 7.2

1. Integrate the function 2𝑥

1+𝑥2 [4 Marks]

Solution:

We need to integrate 2𝑥

1+𝑥2

Let 1 + 𝑥2 = 𝑡

[𝟏

𝟐 Mark]

Differentiating with respect to 𝑡, we get

2𝑥𝑑𝑥

𝑑𝑡= 1

∴ 2𝑥 𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫2𝑥

1+𝑥2 𝑑𝑥 = ∫1

𝑡𝑑𝑡

[𝟏

𝟐 Mark]

= log|𝑡| + 𝐶 [∵ ∫1

𝑥𝑑𝑥 = log |𝑥| + 𝐶 ]

[1 Mark]

= log|1 + 𝑥2| + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]



Hence, integration of the function 2𝑥

1+𝑥2 = log(1 + 𝑥2) + 𝐶

[𝟏

𝟐 Mark]

2. Integrate the function (log 𝑥)2

𝑥 [4 Marks]

Solution:

We need to integrate (log 𝑥)2

𝑥

Let log|𝑥| = 𝑡

[𝟏

𝟐 Mark]


1

𝑥

𝑑𝑥

𝑑𝑡= 1

∴1

𝑥𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫(log|𝑥|)2

𝑥𝑑𝑥 = ∫

(𝑡)2

𝑥𝑑𝑥 [

𝟏

𝟐 Mark]

= ∫(𝑡)2𝑑𝑡

=𝑡3

3+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

=(log|𝑥|)3

3+ 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function (log 𝑥)2

𝑥=

(log|𝑥|)3

3+ 𝐶.

[𝟏

𝟐 Mark]

3. Integrate the function 1

𝑥+𝑥 log 𝑥 [4 Marks]

Solution:

We need to integrate 1

𝑥+𝑥 log 𝑥



=1

𝑥(1 + log 𝑥)

Let 1 + log 𝑥 = 𝑡

[𝟏

𝟐 Mark]


1

𝑥

𝑑𝑥

𝑑𝑡= 1

∴1


⇒ ∫1

𝑥(1+log 𝑥)𝑑𝑥 = ∫

1

𝑡𝑑𝑡 [∵ ∫

1

𝑥𝑑𝑥 = log 𝑥 + 𝐶]

[𝟏

𝟐 Mark]

= log|𝑡| + 𝐶

[𝟏

𝟐 Mark]

= log|1 + log 𝑥| + 𝐶 [substituting 𝑡] [1 Mark]

Hence, integration of the function 1

𝑥+𝑥 log 𝑥 = = log|1 + log 𝑥| + 𝐶

[𝟏

𝟐 Mark]

4. Integrate the function sin 𝑥 sin (cos 𝑥) [4 Marks]

Solution:

We need to integrate sin 𝑥 sin (cos 𝑥)

Let cos 𝑥 = 𝑡

[𝟏

𝟐 Mark]


−sin 𝑥𝑑𝑥

𝑑𝑡= 1

∴ −sin 𝑥 𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫ sin 𝑥 sin(cos 𝑥)𝑑𝑥 = − ∫ sin 𝑡 𝑑𝑡

[𝟏

𝟐 Mark]

= −[− cos 𝑡] + 𝐶 [∵ ∫ sin 𝑥 dx = − 𝑐𝑜𝑠 𝑥 + 𝐶] [1 Mark]



= cos 𝑡 + 𝐶

= cos(cos 𝑥) + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function sin 𝑥 sin (cos 𝑥) = cos(cos 𝑥) + 𝐶

[𝟏

𝟐 Mark]

5. Integrate the function sin(𝑎𝑥 + 𝑏) cos(𝑎𝑥 + 𝑏) [4 Marks]

Solution:

We need to integrate sin(𝑎𝑥 + 𝑏) cos(𝑎𝑥 + 𝑏)

= 2 sin(𝑎𝑥+𝑏) cos(𝑎𝑥+𝑏)

2 [∵ multiplying and dividing it by 2 ]

=sin 2(𝑎𝑥+𝑏)

2 [∵ 2 sin(𝑥) cos(𝑥) = sin 2(𝑥)]

[𝟏

𝟐 Mark]

Let 2(𝑎𝑥 + 𝑏) = 𝑡


2𝑎𝑑𝑥

𝑑𝑡= 1

∴ 2𝑎𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫sin2(𝑎𝑥 + 𝑏)

2𝑑𝑥 =

1

2∫

sin 𝑡 𝑑𝑡

2𝑎

=1

4𝑎[− cos 𝑡] + 𝐶 [∵ ∫ sin 𝑥 dx = − 𝑐𝑜𝑠 𝑥 + 𝐶 ] [1 Mark]

= −1

4𝑎cos 2(𝑎𝑥 + 𝑏) + 𝐶 [substituting 𝑡] [1 Mark]

Hence, integration of the function sin(𝑎𝑥 + 𝑏) cos(𝑎𝑥 + 𝑏) = −1

4𝑎cos 2(𝑎𝑥 + 𝑏) + 𝐶

[𝟏

𝟐 Mark]

6. Integrate the function √𝑎𝑥 + 𝑏 [4 Marks]



Solution:

We need to integrate √𝑎𝑥 + 𝑏

Let 𝑎𝑥 + 𝑏 = 𝑡

[𝟏

𝟐 Mark]


𝑎𝑑𝑥

𝑑𝑡= 1

⇒ 𝑎𝑑𝑥 = 𝑑𝑡

∴ 𝑑𝑥 =1

𝑎𝑑𝑡

[1 Mark]

⇒ ∫(𝑎𝑥 + 𝑏)1

2𝑑𝑥 =1

𝑎∫ 𝑡

1

2𝑑𝑡

[𝟏

𝟐 Mark]

=1

𝑎(

𝑡32

3

2

) + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶]

[1 Mark]

=2

3𝑎(𝑎𝑥 + 𝑏)

3

2 + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function √𝑎𝑥 + 𝑏 =2

3𝑎(𝑎𝑥 + 𝑏)

3

2 + 𝐶

[𝟏

𝟐 Mark]

7. Integrate the function 𝑥√𝑥 + 2 [4 Marks]

Solution:

We need to integrate 𝑥√𝑥 + 2

Let 𝑥 + 2 = 𝑡

[𝟏

𝟐 Mark]


𝑑𝑥

𝑑𝑡= 1



∴ 𝑑𝑥 = 𝑑𝑡

[1 Mark]

⇒ ∫ 𝑥√𝑥 + 2𝑑𝑥 = ∫(𝑡 − 2)√𝑡𝑑𝑡

[𝟏

𝟐 Mark]

= ∫ (𝑡3

2 − 2𝑡1

2) 𝑑𝑡

= ∫ 𝑡32𝑑𝑡 − 2 ∫ 𝑡

12 𝑑𝑡

=𝑡

52

5

2

− 2 (𝑡

32

3

2

) + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶]

[1 Mark]

=2

5𝑡

52 −

4

3𝑡

32 + 𝐶

=2

5(𝑥 + 2)

5

2 −4

3(𝑥 + 2)

3


[𝟏

𝟐 Mark]

Hence, integration of the function, 𝑥√𝑥 + 2 =2

5(𝑥 + 2)

5

2 −4

3(𝑥 + 2)

3

2 + 𝐶

[𝟏

𝟐 Mark]

8. Integrate the function 𝑥√1 + 2𝑥2

[4 Marks]

Solution:

We need to integrate 𝑥√1 + 2𝑥2

Let 1 + 2𝑥2 = 𝑡

[𝟏

𝟐 Mark]


4𝑥𝑑𝑥

𝑑𝑡= 1

∴ 4𝑥𝑑𝑥 = 𝑑𝑡

[1 Mark]



⇒ ∫ 𝑥√1 + 2𝑥2 𝑑𝑥 = ∫√𝑡𝑑𝑡

4

[𝟏

𝟐 Mark]

=1

4∫ 𝑡

12𝑑𝑡

=1

4(

𝑡32

3

2

) + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶]

[1 Mark]

=1

6(1 + 2𝑥2)

3


[𝟏

𝟐 Mark]

Hence, integration of the function, 𝑥√1 + 2𝑥2 =1

6(1 + 2𝑥2)

3

2 + 𝐶

[𝟏

𝟐 Mark]

9. Integrate the function (4𝑥 + 2)√𝑥2 + 𝑥 + 1 [4 Marks]

Solution:

We need to integrate (4𝑥 + 2)√𝑥2 + 𝑥 + 1

Let 𝑥2 + 𝑥 + 1 = 𝑡

[𝟏

𝟐 Mark]


(2𝑥 + 1)𝑑𝑥

𝑑𝑡= 1

∴ (2𝑥 + 1)𝑑𝑥 = 𝑑𝑡 [1 Mark]

∫(4𝑥 + 2)√𝑥2 + 𝑥 + 1 𝑑𝑥

= ∫ 2√𝑡 𝑑𝑡

[𝟏

𝟐 Mark]

= 2 ∫ √𝑡 𝑑𝑡

= 2 (𝑡

32

3

2

) + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]



=4

3(𝑥2 + 𝑥 + 1)

3


[𝟏

𝟐 Mark]

Hence, integration of the function, (4𝑥 + 2)√𝑥2 + 𝑥 + 1 =4

3(𝑥2 + 𝑥 + 1)

3

2 + 𝐶

[𝟏

𝟐 Mark]


𝑥−√𝑥 [4 Marks]

Solution:


𝑥−√𝑥

=1

√𝑥(√𝑥 − 1)

Let (√𝑥 − 1) = 𝑡

[𝟏

𝟐 Mark]


1

2√𝑥

𝑑𝑥

𝑑𝑡= 1

∴1

2√𝑥𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫1

√𝑥(√𝑥 − 1)𝑑𝑥

= ∫2

𝑡𝑑𝑡

[𝟏

𝟐 Mark]

= 2 log|𝑡| + 𝐶 [∵ ∫1

𝑥𝑑𝑥 = log 𝑥 + 𝐶] [1 Mark]

= 2 log|√𝑥 − 1| + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function, 1

𝑥−√𝑥= 2 log|√𝑥 − 1| + 𝐶

[𝟏

𝟐 Mark]



11. Integrate the function 𝑥

√𝑥+4, 𝑥 > 0 [4 Marks]

Solution:

We need to integrate 𝑥

√𝑥+4

Let 𝑥 + 4 = 𝑡

[𝟏

𝟐 Mark]


𝑑𝑥

𝑑𝑡= 1

∴ 𝑑𝑥 = 𝑑𝑡 [1 Mark]

= ∫𝑥

√𝑥 + 4𝑑𝑥

= ∫(𝑡−4)

√𝑡𝑑𝑡

[𝟏

𝟐 Mark]

= ∫ (√𝑡 −4

√𝑡) 𝑑𝑡

=𝑡

32

3

2

− 4 (𝑡

12

1

2

) + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

=2

3(𝑡)

32 − 8(𝑡)

12 + 𝐶

=2

3𝑡 ∙ 𝑡

12 − 8𝑡

12 + 𝐶

=2

3𝑡

12(𝑡 − 12) + 𝐶

=2

3(𝑥 + 4)

1

2(𝑥 + 4 − 12) + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

=2

3√𝑥 + 4(𝑥 − 8) + 𝐶

Hence, integration of the function 𝑥

√𝑥+4=

2

3√𝑥 + 4(𝑥 − 8) + 𝐶

[𝟏

𝟐 Mark]



12. Integrate the function (𝑥3 − 1)1

3𝑥5 [4 Marks]

Solution:

We need to integrate (𝑥3 − 1)1

3𝑥5

Let 𝑥3 − 1 = 𝑡

[𝟏

𝟐 Mark]


3𝑥2 𝑑𝑥

𝑑𝑡= 1

∴ 3𝑥2𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫(𝑥3 − 1)13𝑥5𝑑𝑥 = ∫(𝑥3 − 1)

13 𝑥3 ∙ 𝑥2𝑑𝑥

= ∫ 𝑡1

3(𝑡 + 1)𝑑𝑡

3

[𝟏

𝟐 Mark]

=1

3[

𝑡73

7

3

+𝑡

43

4

3

] + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

=1

3[3

7𝑡

73 +

3

4𝑡

43] + 𝐶

=1

7(𝑥3 − 1)

7

3 +1

4(𝑥3 − 1)

4


[𝟏

𝟐 Mark]

Hence, integration of the function (𝑥3 − 1)1

3𝑥5 =1

7(𝑥3 − 1)

7

3 +1

4(𝑥3 − 1)

4

3 + 𝐶

[𝟏

𝟐 Mark]

13. Integrate the function 𝑥2

(2+3𝑥3)3 [4 Marks]

Solution:

We need to integrate 𝑥2

(2+3𝑥3)3



Let 2 + 3𝑥3 = 𝑡

[𝟏

𝟐 Mark]


9𝑥2 𝑑𝑥

𝑑𝑡= 1


⇒ ∫𝑥2

(2+3𝑥3)3 𝑑𝑥 =1

9∫

𝑑𝑡

(𝑡)3 [𝟏

𝟐 Mark]

=1

9[

𝑡−2

−2] + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

=−1

18(

1

𝑡2) + 𝐶

=−1

18(2+3𝑥3)2 + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function 𝑥2

(2+3𝑥3)3 =−1

18(2+3𝑥3)2 + 𝐶

[𝟏

𝟐 Mark]


𝑥(log 𝑥)𝑚 , 𝑥 > 0, 𝑚 ≠ 1 [4 Marks]

Solution:


𝑥(log 𝑥)𝑚

Let log 𝑥 = 𝑡

[𝟏

𝟐 Mark]


1

𝑥

𝑑𝑥

𝑑𝑡= 1

1


⇒ ∫1

𝑥(log 𝑥)𝑚𝑑𝑥



= ∫𝑑𝑡

(𝑡)𝑚

[𝟏

𝟐 Mark]

= (𝑡−𝑚+1

1−𝑚) + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

=(log 𝑥)1−𝑚

(1−𝑚)+ 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]


𝑥(log 𝑥)𝑚 =(log 𝑥)1−𝑚

(1−𝑚)+ 𝐶

[𝟏

𝟐 Mark]


9−4𝑥2 [4 Marks]

Solution:


9−4𝑥2

Let 9 − 4𝑥2 = 𝑡

[𝟏

𝟐 Mark]


−8𝑥𝑑𝑥

𝑑𝑡= 1

∴ −8𝑥 𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫𝑥

9 − 4𝑥2𝑑𝑥

=−1

8∫

1

𝑡𝑑𝑡

[𝟏

𝟐 Mark]

=−1

8log|𝑡| + 𝐶 [∵ ∫

1


=−1

8log|9 − 4𝑥2| + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]




9−4𝑥2 =−1

8log|9 − 4𝑥2| + 𝐶

[𝟏

𝟐 Mark]

16. Integrate the function 𝑒2𝑥+3 [4 Marks]

Solution:

We need to integrate 𝑒2𝑥+3

Let 2𝑥 + 3 = 𝑡

[𝟏

𝟐 Mark]


2𝑑𝑥

𝑑𝑡= 1

∴ 2𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫𝑒2𝑥+3𝑑𝑥

=1

2∫ 𝑒𝑡𝑑𝑡

[𝟏

𝟐 Mark]

=1

2(𝑒𝑡) + 𝐶 [∵ ∫ 𝑒𝑥𝑑𝑥 = 𝑒𝑥 + 𝐶] [1 Mark]

=1

2𝑒(2𝑥+3) + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function 𝑒2𝑥+3 =1

2𝑒(2𝑥+3) + 𝐶

[𝟏

𝟐 Mark]


𝑒𝑥2 [4 Marks]

Solution:


𝑒𝑥2



Let 𝑥2 = 𝑡

[𝟏

𝟐 Mark]


2𝑥𝑑𝑥

𝑑𝑡= 1

∴ 2𝑥𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫𝑥

𝑒𝑥2 𝑑𝑥 =1

2∫

1

𝑒𝑡 𝑑𝑡

[𝟏

𝟐 Mark]

=1

2∫ 𝑒−𝑡 𝑑𝑡

=1

2(

𝑒−𝑡

−1) + 𝐶 [∵ ∫ 𝑒𝑎𝑥𝑑𝑥 =

𝑒𝑎𝑥

𝑎+ 𝐶] [1 Mark]

= −1

2𝑒−𝑥2

+ 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

=−1

2𝑒𝑥2 + 𝐶


𝑒𝑥2 =−1

2𝑒𝑥2 + 𝐶

[𝟏

𝟐 Mark]

18. Integrate the function 𝑒tan−1 𝑥

1+𝑥2 [4 Marks]

Solution:

We need to integrate 𝑒tan−1 𝑥

1+𝑥2

Let tan−1 𝑥 = 𝑡 [∵ 𝑑

𝑑𝑥(𝑡𝑎𝑛−1 𝑥 ) =

1

1+𝑥2 ]


1

1+𝑥2

𝑑𝑥

𝑑𝑡= 1

∴1

1+𝑥2 𝑑𝑥 = 𝑑𝑡 [1 Mark]



⇒ ∫𝑒tan−1 𝑥

1 + 𝑥2𝑑𝑥

= ∫ 𝑒𝑡𝑑𝑡

[𝟏

𝟐 Mark]

= 𝑒𝑡 + 𝐶 [∵ ∫ 𝑒𝑎𝑥𝑑𝑥 = 𝑒𝑎𝑥

𝑎+ 𝐶] [1 Mark]

= 𝑒tan−1 𝑥 + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function 𝑒tan−1 𝑥

1+𝑥2 = 𝑒tan−1 𝑥 + 𝐶

[𝟏

𝟐 Mark]

19. Integrate the function 𝑒2𝑥−1

𝑒2𝑥+1 [4 Marks]

Solution:

We need to integrate 𝑒2𝑥−1

𝑒2𝑥+1

Dividing numerator and denominator by 𝑒𝑥, we obtain

(𝑒2𝑥 − 1)𝑒𝑥

(𝑒2𝑥 + 1)𝑒𝑥

=𝑒𝑥 − 𝑒−𝑥

𝑒𝑥 + 𝑒−𝑥

Let 𝑒𝑥 + 𝑒−𝑥 = 𝑡

[𝟏

𝟐 Mark]


(𝑒𝑥 − 𝑒−𝑥)𝑑𝑥

𝑑𝑡= 1

∴ (𝑒𝑥 − 𝑒−𝑥)𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫𝑒2𝑥 − 1

𝑒2𝑥 + 1𝑑𝑥 = ∫

𝑒𝑥 − 𝑒−𝑥

𝑒𝑥 + 𝑒−𝑥𝑑𝑥

= ∫𝑑𝑡

𝑡

[𝟏

𝟐 Mark]



= log|𝑡| + 𝐶 [∵ ∫1


= log|𝑒𝑥 + 𝑒−𝑥| + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function 𝑒2𝑥−1

𝑒2𝑥+1= log|𝑒𝑥 + 𝑒−𝑥| + 𝐶

[𝟏

𝟐 Mark]

20. Integrate the function 𝑒2𝑥−𝑒−2𝑥

𝑒2𝑥+𝑒−2𝑥 [4 Marks]

Solution:

We need to integrate 𝑒2𝑥−𝑒−2𝑥

𝑒2𝑥+𝑒−2𝑥

Let 𝑒2𝑥 + 𝑒−2𝑥 = 𝑡

[𝟏

𝟐 Mark]


(2𝑒2𝑥 − 2𝑒2𝑥)𝑑𝑥

𝑑𝑡= 1

∴ (2𝑒2𝑥 − 2𝑒2𝑥)𝑑𝑥 = 𝑑𝑡

⇒ 2(𝑒2𝑥 − 𝑒−2𝑥)𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫ (𝑒2𝑥−𝑒−2𝑥

𝑒2𝑥+𝑒−2𝑥 ) 𝑑𝑥 = ∫𝑑𝑡

2𝑡

=1

2∫

1

𝑡𝑑𝑡

[𝟏

𝟐 Mark]

=1

2log|𝑡| + 𝐶 [∵ ∫

1


=1

2log|𝑒2𝑥 + 𝑒−2𝑥| + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function 𝑒2𝑥−𝑒−2𝑥

𝑒2𝑥+𝑒−2𝑥 =1

2log|𝑒2𝑥 + 𝑒−2𝑥| + 𝐶 [

𝟏

𝟐 Mark]



21. Integrate the function tan2(2𝑥 − 3) [4 Marks]

Solution:

We need to integrate tan2(2𝑥 − 3)

= sec2(2𝑥 − 3) − 1

Let 2𝑥 − 3 = 𝑡

[𝟏

𝟐 Mark]


2𝑑𝑥

𝑑𝑡= 1

∴ 2𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫ tan2(2𝑥 − 3)𝑑𝑥 = ∫[(sec2(2𝑥 − 3)) − 1] 𝑑𝑥 [∵ tan2(𝑥)𝑑𝑥 = (sec2(𝑥)) − 1 ]

[𝟏

𝟐 Mark]

=1

2∫(sec2 𝑡)𝑑𝑡 − ∫ 1𝑑𝑥

=1

2∫ sec2 𝑡 𝑑𝑡 − ∫ 1𝑑𝑥

=1

2tan 𝑡 − 𝑥 + 𝐶 [∵ ∫ sec2x 𝑑𝑥 = tan 𝑥 + 𝐶] [1 Mark]

=1

2tan(2𝑥 − 3) − 𝑥 + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function tan2(2𝑥 − 3) =1

2tan(2𝑥 − 3) − 𝑥 + 𝐶

[𝟏

𝟐 Mark]

22. Integrate the function 𝑠𝑒𝑐2(7 − 4𝑥) [4 Marks]

Solution:

We need to integrate 𝑠𝑒𝑐2(7 − 4𝑥)

Let 7 − 4𝑥 = 𝑡

[𝟏

𝟐 Mark]




−4𝑑𝑥

𝑑𝑡= 1

∴ −4𝑑𝑥 = 𝑑𝑡 [1 Mark]

∴ ∫ sec2(7 − 4𝑥)𝑑𝑥 =−1

4∫ sec2 𝑡 𝑑𝑡

[𝟏

𝟐 Mark]

=−1

4(tan 𝑡) + 𝐶 [∵ ∫ sec2x 𝑑𝑥 = tan 𝑥 + 𝐶] [1 Mark]

=−1

4tan(7 − 4𝑥) + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function sec2(7 − 4𝑥) =−1

4tan(7 − 4𝑥) + 𝐶

[𝟏

𝟐 Mark]

23. Integrate the function sin−1 𝑥

√1−𝑥2 [4 Marks]

Solution:

We need to integrate sin−1 𝑥

√1−𝑥2

Let sin−1 𝑥 = 𝑡

[𝟏

𝟐 Mark]


1

√1−𝑥2

𝑑𝑥

𝑑𝑡= 1 [∵

𝑑

𝑑𝑥(𝑠𝑖𝑛−1 𝑥 ) =

1

√1−𝑥2 ]

⇒1

√1−𝑥2𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫sin−1 𝑥

√1−𝑥2𝑑𝑥 = ∫ 𝑡 𝑑𝑡

[𝟏

𝟐 Mark]

=𝑡2

2+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]



=(sin−1 𝑥)

2


[𝟏

𝟐 Mark]

Hence, integration of the function sin−1 𝑥

√1−𝑥2=

(sin−1 𝑥)2

2+ 𝐶

[𝟏

𝟐 Mark]

24. Integrate the function 2 cos 𝑥−3 sin 𝑥

6 cos 𝑥+4 sin 𝑥 [4 Marks]

Solution:

We need to integrate 2 cos 𝑥−3 sin 𝑥

6 cos 𝑥+4 sin 𝑥

= 2 cos 𝑥 − 3 sin 𝑥

6 cos 𝑥 + 4 sin 𝑥

=2 cos 𝑥 − 3 sin 𝑥

2(3 cos 𝑥 + 2 sin 𝑥)

Let 3 cos 𝑥 + 2 sin 𝑥 = 𝑡 [𝟏

𝟐 Mark]


(−3 sin 𝑥 + 2 cos 𝑥)𝑑𝑥

𝑑𝑡= 1 [∵

𝑑

𝑑𝑥(𝑠𝑖𝑛 𝑥 ) = cos 𝑥 𝑎𝑛𝑑

𝑑

𝑑𝑥(𝑐𝑜𝑠 𝑥 ) = − sin 𝑥 ]

⇒ (−3 sin 𝑥 + 2 cos 𝑥)𝑑𝑥 = 𝑑𝑡 [1 Mark]

∫2 cos 𝑥 − 3 sin 𝑥

6 cos 𝑥 + 4 sin 𝑥 𝑑𝑥

= ∫𝑑𝑡

2𝑡

[𝟏

𝟐 Mark]

=1

2∫

1

𝑡𝑑𝑡

=1

2log|𝑡| + 𝐶 [∵ ∫

1


1

2log|2 sin 𝑥 + 3 cos 𝑥| + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]



Hence, integration of the function 2 cos 𝑥−3 sin 𝑥

6 cos 𝑥+4 sin 𝑥 =

1

2log|2 sin 𝑥 + 3 cos 𝑥| + 𝐶.

[𝟏

𝟐 Mark]


cos2𝑥(1−tan 𝑥)2 [4 Marks]

Solution:


cos2 𝑥(1−tan 𝑥)2

=sec2 𝑥

(1−tan 𝑥)2 [∵1

cos2 𝑥 = sec2 𝑥]

Let (1 − tan 𝑥) = 𝑡

[𝟏

𝟐 Mark]


− sec2 𝑥𝑑𝑥

𝑑𝑡= 1 [∵

𝑑

𝑑𝑥tan 𝑥 = sec2 𝑥]

− sec2 𝑥 𝑑𝑥 = 𝑑𝑡 [1 Mark]

⇒ ∫sec2 𝑥

(1−tan 𝑥)2 𝑑𝑥 = ∫−𝑑𝑡

𝑡2 [𝟏

𝟐 Mark]

= − ∫ 𝑡−2𝑑𝑡

= +1

𝑡+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

=1

1−tan 𝑥+ 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]


cos2 𝑥(1−tan 𝑥)2 =1

1−tan 𝑥+ 𝐶 [

𝟏

𝟐 Mark]

26. Integrate the function cos √𝑥

√𝑥 [4 Marks]

Solution:



We need to integrate cos √𝑥

√𝑥

Let √𝑥 = 𝑡

[𝟏

𝟐 Mark]


1

2√𝑥𝑑𝑥 = 𝑑𝑡 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

⇒ ∫cos √𝑥

√𝑥𝑑𝑥 = 2 ∫ cos 𝑡𝑑𝑡

[𝟏

𝟐 Mark]

= 2 sin 𝑡 + 𝐶 [∵ ∫ cos 𝑥 dx = 𝑠𝑖𝑛 𝑥 + 𝐶] [1 Mark]

= 2 sin √𝑥 + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function cos √𝑥

√𝑥= 2 sin √𝑥 + 𝐶

[𝟏

𝟐 Mark]

27. Integrate the function √sin 2𝑥 cos 2𝑥 [4 Marks]

Solution:

We need to integrate √sin 2𝑥 cos 2𝑥

Let sin 2𝑥 = 𝑡

[𝟏

𝟐 Mark]


So, 2 cos 2𝑥 𝑑𝑥 = 𝑑𝑡 [∵ 𝑑

𝑑𝑥(𝑠𝑖𝑛 𝑥 ) = cos 𝑥] [1 Mark]

⇒ ∫ √sin 2𝑥 cos 2𝑥 𝑑𝑥 =1

2∫ √𝑡𝑑𝑡

[𝟏

𝟐 Mark]

=1

2(

𝑡3

23

2

) + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

=1

3𝑡

3

2 + 𝐶



=1

3(sin 2𝑥)

3


[𝟏

𝟐 Mark]

Hence, integration of the function √sin 2𝑥 cos 2𝑥 =1

3(sin 2𝑥)

3

2 + 𝐶

[𝟏

𝟐 Mark]

28. Integrate the function cos 𝑥

√1+sin 𝑥 [4 Marks]

Solution:

We need to integrate cos 𝑥

√1+sin 𝑥

Let 1 + sin 𝑥 = 𝑡

[𝟏

𝟐 Mark]


∴ cos 𝑥𝑑𝑥 = 𝑑𝑡 [∵ 𝑑

𝑑𝑥(𝑠𝑖𝑛 𝑥 ) = cos 𝑥] [1 Mark]

⇒ ∫cos 𝑥

√1+sin 𝑥𝑑𝑥

= ∫𝑑𝑡

√𝑡

[𝟏

𝟐 Mark]

=𝑡

12

1

2

+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

= 2√𝑡 + 𝐶

= 2√1 + sin 𝑥 + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function cos 𝑥

√1+sin 𝑥= 2√1 + sin 𝑥 + 𝐶

[𝟏

𝟐 Mark]

29. Integrate the function cot 𝑥 log sin 𝑥 [4 Marks]



Solution:

We need to integrate cot 𝑥 log sin 𝑥

Let log sin 𝑥 = 𝑡

[𝟏

𝟐 Mark]


⇒1

sin 𝑥∙ cos 𝑥 𝑑𝑥 = 𝑑𝑡 [∵

𝑑

𝑑𝑥(log 𝑥 ) =

1

𝑥 ]

∴ cot 𝑥 𝑑𝑥 = 𝑑𝑡 [∵ cos 𝑥

sin 𝑥= cot 𝑥 ] [1

Mark]

⇒ ∫ cot 𝑥 log sin 𝑥 𝑑𝑥

= ∫ 𝑡 𝑑𝑡

[𝟏

𝟐 Mark]

=𝑡2

2+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶] [1

Mark]

= 1

2(log sin 𝑥)2 + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function cot 𝑥 log sin 𝑥 = 1

2(log sin 𝑥)2 + 𝐶

[𝟏

𝟐 Mark]

30. Integrate the function sin 𝑥

1+cos 𝑥 [4 Marks]

Solution:

We need to integrate sin 𝑥

1+cos 𝑥

Let 1 + cos 𝑥 = 𝑡

[𝟏

𝟐 Mark]




∴ − sin 𝑥 𝑑𝑥 = 𝑑𝑡 [∵ 𝑑

𝑑𝑥(𝑐𝑜𝑠 𝑥 ) = − sin 𝑥 ] [1 Mark]

⇒ ∫sin 𝑥

1+cos 𝑥𝑑𝑥 = ∫ −

𝑑𝑡

𝑡

[𝟏

𝟐 Mark]

= − log|𝑡| + 𝐶 [∵ ∫1


= −log|1 + cos 𝑥| + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function sin 𝑥

1+cos 𝑥= −log|1 + cos 𝑥| + 𝐶

[𝟏

𝟐 Mark]


(1+cos 𝑥)2 [4 Marks]

Solution:


(1+cos 𝑥)2

Let 1 + cos 𝑥 = 𝑡

[𝟏

𝟐 Mark]


− sin 𝑥 𝑑𝑥 = 𝑑𝑡

𝑑𝑥 = −1

sin 𝑥 𝑑𝑡 [1 Mark]

∴ ∫sin 𝑥

(1 + cos 𝑥)2𝑑𝑥

= ∫ −𝑑𝑡

𝑡2

= − ∫ 𝑡−2𝑑𝑡

[𝟏

𝟐 Mark]

=1

𝑡+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

=1

1+cos 𝑥+ 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]



Hence, integration of the function sin 𝑥

(1+cos 𝑥)2 =1

1+cos 𝑥+ 𝐶

[𝟏

𝟐 Mark]


1+cot 𝑥 [6 Marks]

Solution:


1+cot 𝑥

Let 𝐼 = ∫1

1+cot 𝑥𝑑𝑥

= ∫1

1+cos 𝑥

sin 𝑥

𝑑𝑥 [∵ cos 𝑥

sin 𝑥 = cot 𝑥 ]

[𝟏

𝟐 Mark]

= ∫sin 𝑥

sin 𝑥 + cos 𝑥𝑑𝑥

=1

2∫

2 sin 𝑥

sin 𝑥+cos 𝑥𝑑𝑥 [∵ multiplying and dividing by 2 ]

=1

2∫

(sin 𝑥+cos 𝑥 )+(sin 𝑥−cos 𝑥)

(sin 𝑥+cos 𝑥)𝑑𝑥 [∵ adding and subtracting by cos 𝑥 ]

[1 Mark]

=1

2∫ 1𝑑𝑥 +

1

2∫

sin 𝑥−cos 𝑥

sin 𝑥+cos 𝑥𝑑𝑥

=1

2(𝑥) +

1

2∫

sin 𝑥−cos 𝑥

sin 𝑥+cos 𝑥𝑑𝑥

[𝟏

𝟐 Mark]

Let sin 𝑥 + cos 𝑥 = 𝑡

[𝟏

𝟐 Mark]


⇒ (cos 𝑥 − sin 𝑥)𝑑𝑥 = 𝑑𝑡

[1 Mark]

Hence, 𝐼 =𝑥

2+

1

2∫

(sin 𝑥−cos 𝑥)

(cos 𝑥+sin 𝑥)(cos 𝑥−sin 𝑥) (𝑑𝑡)

∴ 𝐼 =𝑥

2+

1

2∫

−(𝑑𝑡)

𝑡

[𝟏

𝟐 Mark]



=𝑥

2−

1

2log|𝑡| + 𝐶 [∵ ∫

1


[1 Mark]

=𝑥

2−

1

2log|sin 𝑥 + cos 𝑥| + 𝐶 [substituting t ]

[𝟏

𝟐 Mark]


1+cot 𝑥=

𝑥

2−

1

2log|sin 𝑥 + cos 𝑥| + 𝐶

[𝟏

𝟐 Mark]


1−tan 𝑥 [6 Marks]

Solution:


1−tan 𝑥

Let 𝐼 = ∫1

1−tan 𝑥 𝑑𝑥

= ∫1

1−sin 𝑥

cos 𝑥

𝑑𝑥 [∵ tan 𝑥 =sin 𝑥

cos 𝑥 ]

[𝟏

𝟐 Mark]

= ∫cos 𝑥

cos 𝑥 − sin 𝑥𝑑𝑥

=1

2∫

2 cos 𝑥

cos 𝑥−sin 𝑥𝑑𝑥 [∵ multiplying and dividing by 2 ]

=1

2∫

(cos 𝑥−sin 𝑥)+(cos 𝑥+sin 𝑥)

(cos 𝑥−sin 𝑥)𝑑𝑥 [∵ adding and subtracting by sin 𝑥 ]

[1 Mark]

=1

2∫ 1𝑑𝑥 +

1

2∫

cos 𝑥 + sin 𝑥

cos 𝑥 + sin 𝑥𝑑𝑥

=𝑥

2+

1

2∫

cos 𝑥+sin 𝑥

cos 𝑥−sin 𝑥𝑑𝑥

[𝟏

𝟐 Mark]

Put cos 𝑥 − sin 𝑥 = 𝑡

[𝟏

𝟐 Mark]




⇒ (− sin 𝑥 − cos 𝑥)𝑑𝑥 = 𝑑𝑡 [∵ 𝑑

𝑑𝑥(𝑠𝑖𝑛 𝑥 ) = cos 𝑥 𝑎𝑛𝑑

𝑑

𝑑𝑥(𝑐𝑜𝑠 𝑥 ) = −sin 𝑥 ]

[1 Mark]

∴ 𝐼 =𝑥

2+

1

2∫

−(𝑑𝑡)

𝑡

[𝟏

𝟐 Mark]

=𝑥

2−

1

2log|𝑡| + 𝐶 [∵ ∫

1


[1 Mark]

=𝑥

2−

1

2log|cos 𝑥 − sin 𝑥| + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]


1−tan 𝑥 =

𝑥

2−

1

2log|cos 𝑥 − sin 𝑥| + 𝐶

[𝟏

𝟐 Mark]

34. Integrate the function √tan 𝑥

sin 𝑥 cos 𝑥 [6 Marks]

Solution:

We need to integrate √tan 𝑥

sin 𝑥 cos 𝑥

Let 𝐼 = ∫√tan 𝑥

sin 𝑥 cos 𝑥 𝑑𝑥

=∫ √tan 𝑥× cos 𝑥

sin 𝑥 cos 𝑥×cos 𝑥𝑑𝑥 [∵ multiplying and dividing by cos 𝑥 ] [1 Mark]

= ∫√tan 𝑥

tan 𝑥 cos2 𝑥 𝑑𝑥 [∵

sin 𝑥

cos 𝑥= tan 𝑥 ]

= ∫sec2 𝑥𝑑𝑥

√tan 𝑥 [∵

1

cos2 𝑥 = sec2 ] [1 Mark]

Let tan 𝑥 = 𝑡

[𝟏

𝟐 Mark]


⇒ sec2 𝑥𝑑𝑥 = 𝑑𝑡 [1 Mark]



∴ 𝐼 = ∫𝑑𝑡

√𝑡

[𝟏

𝟐 Mark]

= 2√𝑡 + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

= 2√tan 𝑥 + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function √tan 𝑥

sin 𝑥 cos 𝑥 = 2√tan 𝑥 + 𝐶

[𝟏

𝟐 Mark]

35. Integrate the function (1+log 𝑥)2

𝑥 [4 Marks]

Solution:

We need to integrate (1+log 𝑥)2

𝑥

Let, 1 + log 𝑥 = 𝑡

[𝟏

𝟐 Mark]


1

𝑥𝑑𝑥 = 𝑑𝑡 [∵

𝑑 (log 𝑥 ])

𝑑𝑥=

1

𝑥]

∴1


⇒ ∫(1+log 𝑥)2

𝑥𝑑𝑥 = ∫ 𝑡2𝑑𝑡

[𝟏

𝟐 Mark]

=𝑡3

3+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶] [1 Mark]

=(1+log 𝑥)3


[𝟏

𝟐 Mark]

Hence, integration of the function (1+log 𝑥)2

𝑥=

(1+log 𝑥)3

3+ 𝐶

[𝟏

𝟐 Mark]



36. Integrate the function (𝑥+1)(𝑥+log 𝑥)2

𝑥 [4 Marks]

Solution:

We need to integrate (𝑥+1)(𝑥+log 𝑥)2

𝑥

= (𝑥+1

𝑥) (𝑥 + log 𝑥)2 = (1 +

1

𝑥) (𝑥 + log 𝑥)2

Let, 𝑥 + log 𝑥 = 𝑡

[𝟏

𝟐 Mark]


∴ (1 +1

𝑥) 𝑑𝑥 = 𝑑𝑡 [∵

𝑑 (log 𝑥)

𝑑𝑥=

1

𝑥] [1 Mark]

⇒ ∫ (1 +1

𝑥) (𝑥 + log 𝑥)2𝑑𝑥

= ∫ 𝑡2𝑑𝑡

[𝟏

𝟐 Mark]

=𝑡3

3+ 𝐶 [1 Mark]

=1

3(𝑥 + log 𝑥)3 + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function (𝑥+1)(𝑥+log 𝑥)2

𝑥 =

1

3(𝑥 + log 𝑥)3 + 𝐶

[𝟏

𝟐 Mark]

37. Integrate the function 𝑥3𝑠𝑖𝑛(tan−1 𝑥4)

1+𝑥8 [6 Marks]

Solution:

We need to integrate 𝑥3𝑠𝑖𝑛(tan−1 𝑥4)

1+𝑥8

Let 𝑥4 = 𝑡 [∵ 𝑑

𝑑𝑥(𝑥𝑛) = 𝑛 𝑥𝑛−1 ] [

𝟏

𝟐 Mark]





⇒ ∫𝑥3(tan−1 𝑥4)

1+𝑥8 𝑑𝑥 =1

4∫

sin(tan−1 𝑡)

1+𝑡2 𝑑𝑡 … (1) [1 Mark]

Let tan−1 𝑡 = 𝑢

[𝟏

𝟐 Mark]

∴1

1+𝑡2 𝑑𝑡 = 𝑑𝑢 [∵ 𝑑

𝑑𝑥(tan−1) =

1

1+𝑥2 ]

[𝟏

𝟐 Mark]

From (1), we obtain

∫𝑥3 sin(tan−1 𝑥4)𝑑𝑥

1 + 𝑥8=

1

4∫ sin 𝑢𝑑𝑢

=1

4(− cos 𝑢) + 𝐶 [∵ ∫ sin 𝑥 dx = − 𝑐𝑜𝑠 𝑥 + 𝐶] [1 Mark]

=−1

4cos(tan−1 𝑡) + 𝐶 [substituting 𝑢]

[𝟏

𝟐 Mark]

=−1

4cos(tan−1 𝑥4) + 𝐶 [substituting 𝑡]

[𝟏

𝟐 Mark]

Hence, integration of the function 𝑥3𝑠𝑖𝑛(tan−1 𝑥4)

1+𝑥8 =−1

4cos(tan−1 𝑥4) + 𝐶

[𝟏

𝟐 Mark]

38. ∫10𝑥9+10𝑥 loge 10

𝑥10+10𝑥 𝑑𝑥 equals [4 Marks]

(A) 10𝑥 − 𝑥10 + C

(B) 10𝑥 + 𝑥10 + C

(C) (10𝑥 − 𝑥10)−1 + C

(D) log(10𝑥 + 𝑥10) + C

Solution:

Let 𝑥10 + 10𝑥 = 𝑡

[𝟏

𝟐 Mark]




∴ (10𝑥9 + 10𝑥 loge 10)𝑑𝑥 = 𝑑𝑡 [∵ 𝑑

𝑑𝑥(𝑥𝑛) = 𝑛 𝑥𝑛−1 ]

[1 Mark]

⇒ ∫10𝑥9+10𝑥 loge 10

𝑥10+10𝑥 𝑑𝑥

= ∫𝑑𝑡

𝑡

[𝟏

𝟐 Mark]

= log 𝑡 + 𝐶 [∵ ∫1


[1 Mark]

= log(10𝑥 + 𝑥10) + 𝐶

[𝟏

𝟐 Mark]

Hence, the correct Answer is D [𝟏

𝟐 Mark]

39. ∫𝑑𝑥

sin2 𝑥 cos2 𝑥 equals [4 Marks]

(A) tan 𝑥 + cot 𝑥 + 𝐶

(B) tan 𝑥 – cot 𝑥 + 𝐶

(C) tan 𝑥 cot 𝑥 + 𝐶

(D) tan 𝑥 – cot 2𝑥 + 𝐶

Solution:

Let 𝐼 = ∫𝑑𝑥

sin2 𝑥 cos2 𝑥

= ∫1

sin2 𝑥 cos2 𝑥𝑑𝑥

= ∫sin2 𝑥+cos2 𝑥

sin2 𝑥 cos2 𝑥 [1 Mark]

= ∫sin2 𝑥

sin2 𝑥 cos2 𝑥𝑑𝑥 + ∫

cos2 𝑥

sin2 𝑥 cos2 𝑥𝑑𝑥 [

𝟏

𝟐 Mark]

= ∫ sec2 𝑥 𝑑𝑥 + ∫ 𝑐𝑜𝑠𝑒𝑐2𝑥𝑑𝑥 [∵ (1

cos2 𝑥) = sec2 𝑥 𝑎𝑛𝑑 (

1

sin2 𝑥) = cosec2 𝑥 ]

[1 Mark]

= tan 𝑥 − cot 𝑥 + 𝐶 [∵ ∫ sec2 𝑥 𝑑𝑥 = tan 𝑥 + 𝑐 𝑎𝑛𝑑 ∫ cosec2 𝑥 𝑑𝑥 = −cot 𝑥 + 𝐶]



[1 Mark]

Hence, the correct Answer is B. [𝟏

𝟐 Mark]

Exercise 7.3

1. Find the integral of the function sin2(2𝑥 + 5) [4

Marks]

Solution:

We need to integrate sin2(2𝑥 + 5)

=1−cos 2(2𝑥+5)

2=

1−cos(4𝑥+10)

2 [∵ sin2(𝑥) =

1−cos(2𝑥)

2 ] [

𝟏

𝟐

Mark]

⇒ ∫ sin2(2𝑥 + 5)𝑑𝑥

= ∫1−cos(4𝑥+10)

2𝑑𝑥

[𝟏

𝟐 Mark]

=1

2∫ 1 𝑑𝑥 −

1

2∫ cos(4𝑥 + 10)𝑑𝑥

=1

2𝑥 −

1

2(

sin(4𝑥+10)

4) + 𝐶 [∵ ∫ cos 𝑥 dx = sin 𝑥 + 𝐶] [2

Marks]

=1

2𝑥 −

1

8sin(4𝑥 + 10) + 𝐶

Hence, integration of the function sin2(2𝑥 + 5) = 1

2𝑥 −

1

8sin(4𝑥 + 10) + 𝐶 [1

Mark]

2. Find the integral of the function sin 3𝑥 cos 4𝑥 [4

Marks]

Solution:

We need to integrate sin 3𝑥 cos 4𝑥

It is known that, sin 𝐴 cos 𝐵 =1

2{sin(𝐴 + 𝐵) + sin(𝐴 − 𝐵)}



∴ ∫ sin 3𝑥 cos 4𝑥 𝑑𝑥 =1

2∫{sin(3𝑥 + 4𝑥) + sin(3𝑥 − 4𝑥)}𝑑𝑥 [1

Mark]

=1

2∫{sin 7𝑥 + sin(−𝑥)}𝑑𝑥

=1

2∫{sin 7𝑥 − sin 𝑥}𝑑𝑥

=1

2∫ sin 7𝑥 𝑑𝑥 −

1

2∫ sin 𝑥 𝑑𝑥

=1

2(

− cos 7𝑥

7) −

1

2(− cos 𝑥) + 𝐶 [∵ ∫ sin 𝑎𝑥 dx =

− cos 𝑎𝑥

𝑎+ 𝐶] [2

Marks]

=− cos 7𝑥

14+

cos 𝑥

2+ 𝐶

Hence, integration of the function sin 3𝑥 cos 4𝑥 =− cos 7𝑥

14+

cos 𝑥

2+ 𝐶 [1

Mark]

3. Find the integral of the function cos 2𝑥 cos 4𝑥 cos 6𝑥 [4

Marks]

Solution:

We need to integrate cos 2𝑥 cos 4𝑥 cos 6𝑥

It is known that, cos 𝐴 cos 𝐵 =1

2{cos(𝐴 + 𝐵) + cos(𝐴 − 𝐵)}

∴ ∫ cos 2𝑥(cos 4𝑥 cos 6𝑥)𝑑𝑥

= ∫ cos 2𝑥 [1

2{cos(4𝑥 + 6𝑥) + cos(4𝑥 − 6𝑥)}] 𝑑𝑥 [1

Mark]

=1

2∫{cos 2𝑥 cos 10𝑥 + cos 2𝑥 cos(−2𝑥)}𝑑𝑥

=1

2∫{cos 2𝑥 cos 10𝑥 + cos 2𝑥 cos2 2𝑥 }𝑑𝑥

=1

2∫ [{

1

2cos(2𝑥 + 10𝑥) + cos(2𝑥 − 10𝑥)} + (

1+cos 4𝑥

2)] 𝑑𝑥 [

𝟏

𝟐

Mark]



=1

4[

sin 12𝑥

12+

sin 8𝑥

8+ 𝑥 +

sin 4𝑥

4] + 𝐶 [∵ ∫ sin 𝑎𝑥 dx =

− cos 𝑎𝑥

𝑎+ 𝐶] [2

Marks]

Hence, integration of the function cos 2𝑥 cos 4𝑥 cos 6𝑥 = 1

4[

sin 12𝑥

12+

sin 8𝑥

8+ 𝑥 +

sin 4𝑥

4] + 𝐶

[𝟏

𝟐 Mark]

4. Find the integral of the function sin3(2𝑥 + 1) [6

Marks]

Solution:

We need to integrate sin3(2𝑥 + 1)

Let 𝐼 = ∫ sin3(2𝑥 + 1)

⇒ ∫ sin3(2𝑥 + 1) 𝑑𝑥 = ∫ sin2(2𝑥 + 1) ∙ sin(2𝑥 + 1)𝑑𝑥

= ∫(1 − cos2(2𝑥 + 1)) sin(2𝑥 + 1)𝑑𝑥 [∵ sin2(𝑥) + cos2(𝑥) = 1 ] [1

Mark]

Let cos(2𝑥 + 1) = 𝑡 [𝟏

𝟐

Mark]

Differentiating with respect to 𝑡,

⇒ −2 sin(2𝑥 + 1)𝑑𝑥 = 𝑑𝑡 [∵ 𝑑

𝑑𝑥(𝑐𝑜𝑠 𝑥) = − sin 𝑥 ] [1

Mark]

⇒ sin(2𝑥 + 1)𝑑𝑥 =−𝑑𝑡

2

Now, 𝐼 = ∫(1 − cos2(2𝑥 + 1)) sin(2𝑥 + 1)𝑑𝑥

⇒ 𝐼 =−1

2∫(1 − 𝑡2)𝑑𝑡 [

𝟏

𝟐

Mark]

=−1

2{𝑡 −

𝑡3

3} + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =

𝑥𝑛+1

𝑛+1+ 𝐶] [2

Marks]

=−1

2{cos(2𝑥 + 1) −

cos3(2𝑥+1)

3} + 𝐶 [∵ Substituting 𝑡]



=− cos(2𝑥+1)

2+

cos3(2𝑥+1)

6+ 𝐶

[𝟏

𝟐 Mark]

Hence, integration of the function sin3(2𝑥 + 1) =− cos(2𝑥+1)

2+

cos3(2𝑥+1)

6+ 𝐶 [

𝟏

𝟐

Mark]

5. Find the integral of the function sin3 𝑥 cos3 𝑥 [6

Marks]

Solution:

We need to integrate sin3 𝑥 cos3 𝑥

Let 𝐼 = ∫ sin3 𝑥 cos3 𝑥 𝑑𝑥

= ∫ cos3 𝑥 ∙ sin2 𝑥 ∙ sin 𝑥 ∙ 𝑑𝑥

= ∫ cos3 𝑥(1 − cos2 𝑥) sin 𝑥 ∙ 𝑑𝑥 [∵ sin2(𝑥) + cos2(𝑥) = 1 ] [1

Mark]

Let cos 𝑥 = 𝑡 [𝟏

𝟐

Mark]

Differentiating with respect to 𝑡,

⇒ − sin 𝑥 ∙ 𝑑𝑥 = 𝑑𝑡 [1

Mark]

⇒ 𝐼 = ∫ 𝑡3(1 − 𝑡2)𝑑𝑡 [𝟏

𝟐

Mark]

= − ∫(𝑡3 − 𝑡5)𝑑𝑡

= − {𝑡4

4−

𝑡6

6} + 𝐶 [2

Marks]

= − {cos4 𝑥

4−

cos6 𝑥

6} + 𝐶 [∵ Substituting 𝑡]

=cos6 𝑥

6−

cos4 𝑥

4+ 𝐶

[𝟏

𝟐 Mark]



Hence, integration of the function sin3 𝑥 cos3 𝑥 =cos6 𝑥

6−

cos4 𝑥

4+ 𝐶 [

𝟏

𝟐

Mark]

6. Find the integral of the function sin 𝑥 sin 2𝑥 sin 3𝑥 [6

Marks]

Solution:

We need to integrate sin 𝑥 sin 2𝑥 sin 3𝑥

We know that, sin 𝐴 sin 𝐵 =1

2 {cos(𝐴 − 𝐵) − cos(𝐴 + 𝐵)}

∴ ∫ sin 𝑥 sin 2𝑥 sin 3𝑥 𝑑𝑥

= ∫ [sin 𝑥 ∙1

2{cos(2𝑥 − 3𝑥) − cos(2𝑥 + 3𝑥)}] 𝑑𝑥 [1

Mark]

=1

2∫(sin 𝑥 cos(−𝑥) − sin 𝑥 cos 5𝑥)𝑑𝑥

=1

2∫(sin 𝑥 cos 𝑥 − sin 𝑥 cos 5𝑥)𝑑𝑥

=1

2∫

sin 2𝑥

2𝑑𝑥 −

1

2∫ sin 𝑥 cos 5𝑥 𝑑𝑥 [∵ 2 sin 𝑥 cos 𝑥 = sin 2𝑥 ]

=1

4[

− cos 2𝑥

2] −

1

2∫ {

1

2sin(𝑥 + 5𝑥) + sin(𝑥 − 5𝑥)} 𝑑𝑥 [2

Marks]

[∵ sin 𝐴 cos 𝐵 =1

2 {sin(𝐴 + 𝐵) + sin(𝐴 − 𝐵)}]

=1

4[− cos 2𝑥

2] −

1

2∫ {

1

2sin(6𝑥) − sin(4𝑥)} 𝑑𝑥

=− cos 2𝑥

8−

1

4[

− cos 6𝑥

3+

cos 4𝑥

4] + 𝐶 [∵ ∫ sin 𝑎𝑥 dx =

− cos 𝑎𝑥

𝑎+ 𝐶 ] [2

Marks]

=− cos 2𝑥

8−

1

8[

− cos 6𝑥

3+

cos 4𝑥

2] + 𝐶

=1

8[

cos 6𝑥

3−

cos 4𝑥

2− cos 2𝑥] + 𝐶

[𝟏

𝟐 Mark]

Hence, integration of the function sin 𝑥 sin 2𝑥 sin 3𝑥 =1

8[

cos 6𝑥

3−

cos 4𝑥

2− cos 2𝑥] + 𝐶 [

𝟏

𝟐

Mark]



7. Find the integral of the function sin 4𝑥 sin 8𝑥 [4

Marks]

Solution:

We need to integrate sin 4𝑥 sin 8𝑥

It is known that, sin 𝐴 sin 𝐵 =1

2[cos(𝐴 − 𝐵) − cos(𝐴 + 𝐵)]

∴ ∫ sin 4𝑥 sin 8𝑥 𝑑𝑥 = ∫ {1

2[cos(4𝑥 − 8𝑥) − cos(4𝑥 + 8𝑥)]} 𝑑𝑥 [1

Mark]

=1

2∫(cos(−4𝑥) − cos 12𝑥)𝑑𝑥

=1

2∫(cos 4𝑥 − cos 12𝑥)𝑑𝑥

=1

2[

sin 4𝑥

4−

sin 12𝑥

12] + 𝐶 [∵ ∫ sin 𝑎𝑥 𝑑𝑥 =

− cos 𝑎𝑥

𝑎+ 𝐶] [2

Marks]

Hence, integration of the function sin 4𝑥 sin 8𝑥 =1

2[

sin 4𝑥

4−

sin 12𝑥

12] + 𝐶 [1

Mark]

8. Find the integral of the function 1−cos 𝑥


Solution:

We need to integrate 1−cos 𝑥

1+cos 𝑥

1−cos 𝑥

1+cos 𝑥=

2 sin2𝑥

2

2 cos2𝑥

2

[2 sin2 𝑥

2= 1 − cos 𝑥 and 2 cos2 𝑥

2 = 1 + cos 𝑥]

= tan2𝑥

2

= (sec2 𝑥

2− 1) [

𝟏

𝟐

Mark]

∴ ∫1−cos 𝑥

1+cos 𝑥𝑑𝑥 = ∫ (sec2 𝑥

2− 1) 𝑑𝑥



= [tan

𝑥2

12

− 𝑥] + 𝐶

= 2 tan𝑥

2− 𝑥 + 𝐶 [1

Mark]

Hence, integration of the function 1−cos 𝑥

1+cos 𝑥= 2 tan

𝑥

2− 𝑥 + 𝐶 [

𝟏

𝟐

Mark]

9. Find the integral of the function cos 𝑥


Solution:


1+cos 𝑥

cos 𝑥

1+cos 𝑥=

cos2𝑥

2−sin2𝑥

2

2 cos2𝑥

2

[cos 𝑥 = cos2 𝑥

2− sin2 𝑥

2and cos 𝑥 = 2 cos2 𝑥

2− 1]

=1

2[1 − tan2 𝑥

2] [1

Mark]

∴ ∫cos 𝑥

1+cos 𝑥𝑑𝑥 =

1

2∫ (1 − tan2 𝑥

2) 𝑑𝑥

=1

2∫ (1 − sec2 𝑥

2+ 1) 𝑑𝑥

=1

2∫ (2 − sec2 𝑥

2) 𝑑𝑥

=1

2[2𝑥 −

tan𝑥

21

2

+ 𝐶]

= 𝑥 − tan𝑥

2+ 𝐶 [2

Marks]

Hence, integration of the function cos 𝑥

1+cos 𝑥= 𝑥 − tan

𝑥

2+ 𝐶 [1

Mark]

10. Find the integral of the function sin4 𝑥 [4 Marks]



Solution:

We need to integrate sin4 𝑥

sin4 𝑥 = sin2 𝑥 sin2 𝑥

= (1−cos 2𝑥

2) (

1−cos 2𝑥

2)

=1

4(1 − cos 2𝑥)2

=1

4[1 + cos2 2𝑥 − 2 cos 2𝑥]

=1

4[1 + (

1 + cos 4𝑥

2) − 2 cos 2𝑥]

=1

4[1 +

1

2+

1

2cos 4𝑥 − 2 cos 2𝑥]

=1

4[

3

2+

1

2cos 4𝑥 − 2 cos 2𝑥] [2

Marks]

∴ ∫ sin4 𝑥 𝑑𝑥 =1

4∫ [

3

2+

1

2cos 4𝑥 − 2 cos 2𝑥] 𝑑𝑥

=1

4[3

2𝑥 +

1

2(

sin 4𝑥

4) −

2 sin 2𝑥

2] + 𝐶

=1

8[3𝑥 +

sin 4𝑥

4− 2 sin 2𝑥] + 𝐶

=3𝑥

8−

1

4sin 2𝑥 +

1

32sin 4𝑥 + 𝐶

Hence, integration of the function sin4 𝑥 =3𝑥

8−

1

4sin 2𝑥 +

1

32sin 4𝑥 + 𝐶 [1

Mark]

11. Find the integral of the function cos4 2𝑥 [4 Marks]

Solution:

We need to integrate cos4 2𝑥

cos4 2𝑥 = (cos2 2𝑥)2

= (1 + cos 4𝑥

2)

2



=1

4[1 + cos2 4𝑥 + 2 cos 4𝑥]

=1

4[1 + (

1 + cos 8𝑥

2) + 2 cos 4𝑥]

=1

4[1 +

1

2+

cos 8𝑥

2+ 2 cos 4𝑥]

=1

4[

3

2+

cos 8𝑥

2+ 2 cos 4𝑥 ] [2

Marks]

∴ ∫ cos4 2𝑥 𝑑𝑥 = ∫ (3

8+

cos 8𝑥

8+

cos 4𝑥

2) 𝑑𝑥

=3

8𝑥 +

sin 8𝑥

64+

sin 4𝑥

8+ 𝐶

Hence, integration of the function cos4 2𝑥 =3

8𝑥 +

sin 8𝑥

64+

sin 4𝑥

8+ 𝐶 [2

Marks]

12. Find the integral of the function sin2 𝑥


Solution:

We need to integrate sin2 𝑥

1+cos 𝑥

sin2 𝑥

1+cos 𝑥=

(2 sin𝑥

2cos

𝑥

2)

2

2 cos2𝑥

2

[sin 𝑥 = 2 sin𝑥

2cos

𝑥

2; cos 𝑥 = 2 cos2 𝑥

2− 1]

=4 sin2 𝑥

2 cos2 𝑥2

2 cos2 𝑥2

= 2 sin2𝑥

2

= 1 − cos 𝑥 [2

Marks]

∴ ∫sin2 𝑥

1 + cos 𝑥𝑑𝑥 = ∫(1 − cos 𝑥)𝑑𝑥

= 𝑥 − sin 𝑥 + 𝐶



Hence, integration of the function sin2 𝑥

1+cos 𝑥= 𝑥 − sin 𝑥 + 𝐶 [2

Marks]

13. Find the integral of the function cos 2𝑥−cos 2𝛼

cos 𝑥−cos 𝛼 [4

Marks]

Solution:

We need to integrate cos 2𝑥−cos 2𝛼

cos 𝑥−cos 𝛼

cos 2𝑥−cos 2𝛼

cos 𝑥−cos 𝛼=

−2 sin2𝑥+2𝛼

2sin

2𝑥−2𝛼

2

−2 sin𝑥+𝛼

2sin

𝑥−𝛼

2

[cos 𝐶 − cos 𝐷 = −2 sin𝐶+𝐷

2sin

𝐶−𝐷

2] [1

Mark]

=sin(𝑥 + 𝛼) sin(𝑥 − 𝛼)

sin (𝑥 + 𝛼

2 ) sin (𝑥 − 𝛼

2 )

=[2 sin (

𝑥 + 𝛼2

)] [2 sin (𝑥 − 𝛼

2) cos (

𝑥 − 𝛼2

)]

sin (𝑥 + 𝛼

2) sin (

𝑥 − 𝛼2

)

= 4 cos (𝑥 + 𝛼

2) cos (

𝑥 − 𝛼

2)

= 2 [cos (𝑥 + 𝛼

2+

𝑥 − 𝛼

2) + cos (

𝑥 + 𝛼

2−

𝑥 − 𝛼

2)]

= 2[cos(𝑥) + cos 𝛼]

= 2 cos 𝑥 + 2 cos 𝛼 [2

Marks]

∴ ∫cos 2𝑥 − cos 2𝛼

cos 𝑥 − cos 𝛼𝑑𝑥 = ∫(2 cos 𝑥 + 2 cos 𝛼) 𝑑𝑥

= [2 sin 𝑥 + 2𝑥 cos 𝛼] + 𝐶

= 2[sin 𝑥 + 𝑥 cos 𝛼] + 𝐶

Hence, integration of the function cos 2𝑥−cos 2𝛼

cos 𝑥−cos 𝛼= 2[sin 𝑥 + 𝑥 cos 𝛼] + 𝐶 [1

Mark]



14. Find the integral of the function cos 𝑥−sin 𝑥

1+sin 2𝑥 [4

Marks]

Solution:

We need to integrate cos 𝑥−sin 𝑥

1+sin 2𝑥

cos 𝑥−sin 𝑥

1+sin 2𝑥=

cos 𝑥−sin 𝑥

(sin2 𝑥+cos2 𝑥)+2 sin 𝑥 cos 𝑥

[sin2 𝑥 + cos2 𝑥 = 1; sin 2𝑥 = 2 sin 𝑥 cos 𝑥]

=cos 𝑥−sin 𝑥

(sin 𝑥+cos 𝑥)2 [1

Mark]

Let sin 𝑥 + cos 𝑥 = 𝑡 [𝟏

𝟐

Mark]

∴ (cos 𝑥 − sin 𝑥)𝑑𝑥 = 𝑑𝑡 [1

Mark]

⇒ ∫cos 𝑥 − sin 𝑥

1 + sin 2𝑥𝑑𝑥 = ∫

cos 𝑥 − sin 𝑥

(sin 𝑥 + cos 𝑥)2𝑑𝑥

= ∫𝑑𝑡

𝑡2 [𝟏

𝟐

Mark]

= ∫ 𝑡−2𝑑𝑡

= −𝑡−1 + 𝐶

= −1

𝑡+ 𝐶

=−1

sin 𝑥+cos 𝑥+ 𝐶

Hence, integration of the function cos 𝑥−sin 𝑥

1+sin 2𝑥=

−1

sin 𝑥+cos 𝑥+ 𝐶 [1

Mark]

15. Find the integral of the function tan3 2𝑥 sec 2𝑥 [4 Marks]

Solution:

We need to integrate tan3 2𝑥 sec 2𝑥



tan3 2𝑥 sec 2𝑥 = tan2 2𝑥 tan 2𝑥 sec 2𝑥

= (sec2 2𝑥 − 1) tan 2𝑥 sec 2𝑥

= sec2 2𝑥 ∙ tan 2𝑥 sec 2𝑥 − tan 2𝑥 sec 2𝑥 [1

Mark]

∴ ∫ tan3 2𝑥 sec 2𝑥 𝑑𝑥 = ∫ sec2 2𝑥 tan 2𝑥 sec 2𝑥 𝑑𝑥 − ∫ tan 2𝑥 sec 2𝑥 𝑑𝑥

= ∫ sec2 2𝑥 tan 2𝑥 sec 2𝑥 𝑑𝑥 −sec 2𝑥

2+ 𝐶 [

𝟏

𝟐

Mark]

Let sec 2𝑥 = 𝑡 [𝟏

𝟐

Mark]

∴ 2 sec 2𝑥 tan 2𝑥 𝑑𝑥 = 𝑑𝑡

∴ ∫ tan3 2𝑥 sec 2𝑥 𝑑𝑥 =1

2∫ 𝑡2𝑑𝑡 −

sec 2𝑥

2+ 𝐶 [1

Mark]

=𝑡3

6−

sec 2𝑥

2+ 𝐶

=(sec 2𝑥)3

6−

sec 2𝑥

2+ 𝐶

Hence, integration of the function tan3 2𝑥 sec 2𝑥 =(sec 2𝑥)3

6−

sec 2𝑥

2+ 𝐶 [1

Mark]

16. Find the integral of the function tan4 𝑥 [4 Marks]

Solution:

We need to integrate tan4 𝑥

= tan2 𝑥 ∙ tan2 𝑥

= (sec2 𝑥 − 1)tan2 𝑥

= sec2 𝑥 tan2 𝑥 − tan2 𝑥

= sec2 𝑥 tan2 𝑥 − (sec2 𝑥 − 1)

= sec2 𝑥 tan2 𝑥 − sec2 𝑥 + 1 [1

Mark]

∴ ∫ tan4 𝑥 𝑑𝑥 = ∫ sec2 𝑥 tan2 𝑥 𝑑𝑥 − ∫ sec2 𝑥 𝑑𝑥 + ∫ 1 ∙ 𝑑𝑥



= ∫ sec2 𝑥 tan2 𝑥 𝑑𝑥 − tan 𝑥 + 𝑥 + 𝐶 … (𝑖) [1

Mark]

Consider ∫ sec2 𝑥 tan2 𝑥 𝑑𝑥

Let tan 𝑥 = 𝑡 ⇒ sec2 𝑥 𝑑𝑥 = 𝑑𝑡

⇒ ∫ sec2 𝑥 tan2 𝑥𝑑𝑥 = ∫ 𝑡2 𝑑𝑡 =𝑡3

3=

tan3 𝑥

3 [1

Mark]

From equation (𝑖), we obtain

∫ tan4 𝑥 𝑑𝑥 =tan3 𝑥

3− tan 𝑥 + 𝑥 + 𝐶

Hence, integration of the function tan4𝑥 =1

3tan3 𝑥 − tan 𝑥 + 𝑥 + 𝐶 [1

Mark]

17. Find the integral of the function sin3𝑥+cos3 𝑥

sin2 𝑥 cos2 𝑥 [4

Marks]

Solution:

We need to integrate sin3𝑥+cos3 𝑥

sin2 𝑥 cos2 𝑥

sin3𝑥+cos3 𝑥

sin2 𝑥 cos2 𝑥=

sin3 𝑥

sin2 𝑥 cos2 𝑥+

cos3 𝑥

sin2 cos2 𝑥

=sin 𝑥

cos2 𝑥+

cos 𝑥

sin2 𝑥

= tan 𝑥 sec 𝑥 + cot 𝑥 cosec 𝑥 [2

Marks]

∴ ∫sin3 𝑥 + cos3 𝑥

sin2 𝑥 cos2 𝑥𝑑𝑥 = ∫(tan 𝑥 sec 𝑥 + cot 𝑥 cosec 𝑥) 𝑑𝑥

= sec 𝑥 − cosec 𝑥 + 𝐶

Hence, integration of the function sin3𝑥+cos3 𝑥

sin2 𝑥 cos2 𝑥 = sec 𝑥 − cosec 𝑥 + 𝐶 [2

Marks]

18. Find the integral of the function cos 2𝑥+2 sin2 𝑥

cos2 𝑥 [2 Marks]



Solution:

We need to integrate cos 2𝑥+2 sin2 𝑥

cos2 𝑥

=cos 2𝑥+(1−cos 2𝑥)

cos2 𝑥[cos 2𝑥 = 1 − 2 sin2 𝑥]

=1

cos2 𝑥

= sec2 𝑥 [1

Mark]

Hence, ∫cos 2𝑥+2 sin2 𝑥

cos2 𝑥𝑑𝑥 = ∫ sec2 𝑥 𝑑𝑥 = tan 𝑥 + 𝐶 [1

Mark]

19. Find the integral of the function 1

sin 𝑥 cos3 𝑥 [4

Marks]

Solution:


sin 𝑥 cos3 𝑥

=sin2 𝑥 + cos2 𝑥

sin 𝑥 cos3 𝑥

=sin 𝑥

cos3 𝑥+

1

sin 𝑥 cos 𝑥

= tan 𝑥 sec2 𝑥 +sec2 𝑥

sin 𝑥

cos 𝑥cos2 𝑥 sec2 𝑥

= tan 𝑥 sec2 𝑥 +sec2 𝑥

tan 𝑥 [1

Mark]

∴ ∫1

sin 𝑥 cos3 𝑥𝑑𝑥 = ∫ tan 𝑥 sec2 𝑥 𝑑𝑥 + ∫

sec2 𝑥

tan 𝑥𝑑𝑥

Let tan 𝑥 = 𝑡 ⇒ sec2 𝑥 𝑑𝑥 = 𝑑𝑡

⇒ ∫1

sin 𝑥 cos3 𝑥𝑑𝑥 = ∫ 𝑡𝑑𝑡 + ∫

1

𝑡𝑑𝑡 [1

Mark]

=𝑡2

2+ log|𝑡| + 𝐶



=1

2tan2 𝑥 + log|tan 𝑥| + 𝐶


sin 𝑥 cos3 𝑥=

1

2tan2 𝑥 + log|tan 𝑥| + 𝐶 [2

Marks]

20. Find the integral of the function cos 2𝑥

(cos 𝑥+sin 𝑥)2 [6

Marks]

Solution:

We need to integrate cos 2𝑥

(cos 𝑥+sin 𝑥)2

=cos 2𝑥

cos2 𝑥+sin2 𝑥+2 sin 𝑥 cos 𝑥=

cos 2𝑥

1+sin 2𝑥

∴ ∫cos 2𝑥

(cos 𝑥+sin 𝑥)2 𝑑𝑥 = ∫cos 2𝑥

(1+sin 2𝑥)𝑑𝑥 [2

Marks]

Let 1 + sin 2𝑥 = 𝑡

⇒ 2 cos 2𝑥 𝑑𝑥 = 𝑑𝑡 [1

Mark]

∴ ∫cos 2𝑥

(cos 𝑥+sin 𝑥)2 𝑑𝑥 = ∫cos 2𝑥

(1+sin 2𝑥)𝑑𝑥

=1

2∫

1

𝑡𝑑𝑡 [1

Mark]

=1

2log|𝑡| + 𝐶

=1

2log|1 + sin 2𝑥| + 𝐶

=1

2log|(sin 𝑥 + cos 𝑥)2| + 𝐶

= log|sin 𝑥 + cos 𝑥| + 𝐶

Hence, integration of the function cos 2𝑥

(cos 𝑥+sin 𝑥)2 = log|sin 𝑥 + cos 𝑥| + 𝐶 [2

Marks]

21. Find the integral of the function sin−1(cos 𝑥) [6 Marks]



Solution:

We need to integrate sin−1(cos 𝑥)

Let cos 𝑥 = 𝑡 Hence, sin 𝑥 = √1 − 𝑡2

⇒ (− sin 𝑥)𝑑𝑥 = 𝑑𝑡

𝑑𝑥 =−𝑑𝑡

sin 𝑥

𝑑𝑥 =−𝑑𝑡

√1−𝑡2 [1

Mark]

∴ ∫ sin−1(cos 𝑥)𝑑𝑥 = ∫ sin−1 𝑡 (−𝑑𝑡

√1 − 𝑡2)

= − ∫sin−1 𝑡

1−𝑡2 𝑑𝑡 [1

Mark]

Let sin−1 𝑡 = 𝑢

⇒1

√1−𝑡2𝑑𝑡 = 𝑑𝑢 [1

Mark]

∴ ∫ sin−1(cos 𝑥)𝑑𝑥 = − ∫ 𝑢𝑑𝑢

= −𝑢2

2+ 𝐶

=−(sin−1 𝑡)2

2+ 𝐶

= −[sin−1(cos 𝑥)]

2

2+ 𝐶 … (𝑖) [1

Mark]

It is known that,

sin−1 𝑥 + cos−1 𝑥 =𝜋

2

∴ sin−1(cos 𝑥) =𝜋

2− cos−1(cos 𝑥) = (

𝜋

2− 𝑥) [1

Mark]

Substituting in equation (𝑖), we obtain

∫ sin−1(cos 𝑥)𝑑𝑥 =−[

𝜋

2−𝑥]

2

2+ 𝐶



= −1

2(

𝜋2

4+ 𝑥2 − 𝜋𝑥) + 𝐶

= −𝜋2

8−

𝑥2

2+

1

2𝜋𝑥 + 𝐶

=𝜋𝑥

2−

𝑥2

2+ 𝐶1

Hence, integration of the function sin−1(cos 𝑥) =𝜋𝑥

2−

𝑥2

2+ 𝐶1 [1

Mark]

22. Find the integral of the function 1

cos(𝑥−𝑎) cos(𝑥−𝑏) [6

Marks]

Solution:


cos(𝑥−𝑎) cos(𝑥−𝑏)

=1

sin(𝑎−𝑏)[

sin(𝑎−𝑏)

cos(𝑥−𝑎) cos(𝑥−𝑏)] [1

Mark]

=1

sin(𝑎−𝑏)[

sin[(𝑥−𝑏)−(𝑥−𝑎)]

cos(𝑥−𝑎) cos(𝑥−𝑏)]

=1

sin(𝑎−𝑏)

[sin(𝑥−𝑏) cos(𝑥−𝑎)−cos(𝑥−𝑏) sin(𝑥−𝑎)]

cos(𝑥−𝑎) cos(𝑥−𝑏)

=1

sin(𝑎−𝑏)[tan(𝑥 − 𝑏) − tan(𝑥 − 𝑎)] [2

Marks]

⇒ ∫1

cos(𝑥 − 𝑎) cos(𝑥 − 𝑏)𝑑𝑥 =

1

sin(𝑎 − 𝑏)∫[tan(𝑥 − 𝑏) − tan(𝑥 − 𝑎)]𝑑𝑥

=1

sin(𝑎−𝑏)[− log|cos(𝑥 − 𝑏)| + log |cos(𝑥 − 𝑎)|] + 𝐶 [2

Marks]

=1

sin(𝑎−𝑏)[log |

cos(𝑥−𝑎)

cos(𝑥−𝑏)|] + 𝐶


cos(𝑥−𝑎) cos(𝑥−𝑏)=

1

sin(𝑎−𝑏)[log |

cos(𝑥−𝑎)

cos(𝑥−𝑏)|] + 𝐶 [1

Mark]



23. ∫sin2 𝑥−cos2 𝑥

sin2 𝑥 cos2 𝑥𝑑𝑥 is equal to [2

Marks]

(A) tan 𝑥 + cot 𝑥 + 𝐶

(B) tan 𝑥 + cosec 𝑥 + 𝐶

(C) − tan 𝑥 + cot 𝑥 + 𝐶

(D) tan 𝑥 + sec 𝑥 + 𝐶

Solution:

Given integral is ∫sin2 𝑥−cos2 𝑥

sin2 𝑥 cos2 𝑥𝑑𝑥

= ∫ (sin2 𝑥

sin2 𝑥 cos2 𝑥−

cos2 𝑥

sin2 𝑥 cos2 𝑥) 𝑑𝑥

= ∫(sec2 𝑥 − 𝑐𝑜𝑠𝑒𝑐2𝑥)𝑑𝑥

= tan 𝑥 + cot 𝑥 + 𝐶

Hence, ∫sin2 𝑥−cos2 𝑥

sin2 𝑥 cos2 𝑥𝑑𝑥 = tan 𝑥 + cot 𝑥 + 𝐶

Hence, the correct Answer is A. [2

Marks]

24. ∫𝑒𝑥(1+𝑥)

cos2(𝑒𝑥𝑥)𝑑𝑥 equals [4 Marks]

(A) − cot(𝑒𝑥𝑥) + 𝐶

(B) tan(𝑥𝑒𝑥) + 𝐶

(C) tan(𝑒𝑥) + 𝐶

(D) cot(𝑒𝑥) + 𝐶

Solution:

Given integral is ∫𝑒𝑥(1+𝑥)

cos2(𝑒𝑥𝑥)𝑑𝑥

Let 𝑒𝑥𝑥 = 𝑡

[𝟏

𝟐 Mark]



⇒ (𝑒𝑥 ∙ 𝑥 + 𝑒𝑥 ∙ 1)𝑑𝑥 = 𝑑𝑡

𝑒𝑥(𝑥 + 1)𝑑𝑥 = 𝑑𝑡 [1

Mark]

∴ ∫𝑒𝑥(1+𝑥)

cos2(𝑒𝑥𝑥)𝑑𝑥 = ∫

𝑑𝑡

cos2 𝑡

[𝟏

𝟐 Mark]

= ∫ sec2 𝑡 𝑑𝑡

= tan 𝑡 + 𝐶 [1

Mark]

= tan(𝑒𝑥𝑥) + 𝐶

[𝟏

𝟐 Mark]

Hence, ∫𝑒𝑥(1+𝑥)

cos2(𝑒𝑥𝑥)𝑑𝑥 = tan(𝑒𝑥𝑥) + 𝐶

Hence, the correct Answer is B.

[𝟏

𝟐 Mark]

Exercise 7.4

1. Integrate the function 3𝑥2

𝑥6+1 [2

Marks]

Solution:

We need to integrate the function 3𝑥2

𝑥6+1

Let 𝑥3 = 𝑡

∴ 3𝑥2𝑑𝑥 = 𝑑𝑡

[𝟏

𝟐 Mark]

⇒ ∫3𝑥2

𝑥6 + 1𝑑𝑥 = ∫

𝑑𝑡

𝑡2 + 1

= tan−1 𝑡 + 𝐶

= tan−1(𝑥3) + 𝐶 [Substituting 𝑡] [1

Mark]



Hence, integration of 3𝑥2

𝑥6+1= tan−1(𝑥3) + 𝐶 [

𝟏

𝟐

Mark]


√1+4𝑥2 [2

Marks]

Solution:

We need to integrate the function 1

√1+4𝑥2

Let 2𝑥 = 𝑡

∴ 2𝑑𝑥 = 𝑑𝑡

[𝟏

𝟐 Mark]

⇒ ∫1

√1 + 4𝑥2𝑑𝑥 =

1

2∫

𝑑𝑡

√1 + 𝑡2

=1

2[log|𝑡 + √𝑡2 + 1|] + 𝐶 [∫

1

√𝑥2+𝑎2𝑑𝑥 = log|𝑥 + √𝑥2 + 𝑎2|]

=1

2log|2𝑥 + √4𝑥2 + 1| + 𝐶 [Substituting 𝑡] [1

Mark]

Hence, integration of 1

√1+4𝑥2=

1

2log|2𝑥 + √4𝑥2 + 1| + 𝐶

[𝟏

𝟐 Mark]


√(2−𝑥)2+1 [2 Marks]

Solution:


√(2−𝑥)2+1

Let 2 − 𝑥 = 𝑡

⇒ −𝑑𝑥 = 𝑑𝑡

[𝟏

𝟐 Mark]



⇒ ∫1

√(2 − 𝑥)2 + 1𝑑𝑥 = − ∫

1

√𝑡2 + 1𝑑𝑡

= − log|𝑡 + √𝑡2 + 1| + C [∫1

√𝑥2+𝑎2𝑑𝑥 = log|𝑥 + √𝑥2 + 𝑎2|]

= − log |2 − 𝑥 + √(2 − 𝑥)2 + 1| + C [Substituting 𝑡]

= log |1

(2−𝑥)+√𝑥2−4𝑥+5| + C [1

Mark]


√(2−𝑥)2+1= log |

1

(2−𝑥)+√𝑥2−4𝑥+5| + C

[𝟏

𝟐 Mark]


√9−25𝑥2 [2 Marks]

Solution:


√9−25𝑥2

Let 5𝑥 = 𝑡

∴ 5𝑑𝑥 = 𝑑𝑡 [𝟏

𝟐

Mark]

⇒ ∫1

√9−25𝑥2𝑑𝑥 =

1

5∫

1

√9−𝑡2𝑑𝑡

=1

5∫

1

√32−𝑡2𝑑𝑡

=1

5sin−1 (

𝑡

3) + 𝐶 [Since, ∫

1

√𝑎2−𝑥2𝑑𝑥 = sin−1 𝑥

𝑎]

=1

5sin−1 (

5𝑥

3) + 𝐶 [Substituting 𝑡] [1

Mark]


√9−25𝑥2=

1

5sin−1 (

5𝑥

3) + 𝐶

[𝟏

𝟐 Mark]




1+2𝑥4 [2

Marks]

Solution:

We need to integrate the function 3𝑥

1+2𝑥4

Let √2𝑥2 = 𝑡

∴ 2√2𝑥𝑑𝑥 = 𝑑𝑡

[𝟏

𝟐 Mark]

⇒ ∫3𝑥

1 + 2𝑥4𝑑𝑥 =

3

2√2∫

𝑑𝑡

1 + 𝑡2

=3

2√2[tan−1𝑡] + C

=3

2√2tan−1(√2𝑥2) + C [Substituting 𝑡] [1

Mark]

Hence, integration of 3𝑥

1+2𝑥4 =3

2√2tan−1(√2𝑥2) + C

[𝟏

𝟐 Mark]


1−𝑥6 [2

Marks]

Solution:

We need to integrate the function 𝑥2

1−𝑥6

Let 𝑥3 = 𝑡

∴ 3𝑥2𝑑𝑥 = 𝑑𝑡

[𝟏

𝟐 Mark]

⇒ ∫𝑥2

1 − 𝑥6𝑑𝑥 =

1

3∫

𝑑𝑡

1 − 𝑡2

=1

3[1

2log |

1 + 𝑡

1 − 𝑡|] + C



=1

6log |

1+𝑥3

1−𝑥3| + C [Substituting 𝑡] [1

Mark]

Hence, integration of 𝑥2

1−𝑥6 = 1

6log |

1+𝑥3

1−𝑥3| + C

[𝟏

𝟐 Mark]

7. Integrate the function 𝑥−1

√𝑥2−1 [4

Marks]

Solution:

We need to integrate the function 𝑥−1

√𝑥2−1

∫𝑥−1

√𝑥2−1𝑑𝑥 = ∫

𝑥

√𝑥2−1𝑑𝑥 − ∫

1

√𝑥2−1𝑑𝑥

[𝟏

𝟐 Mark]

For ∫𝑥

√𝑥2−1𝑑𝑥, let 𝑥2 − 1 = 𝑡 ⇒ 2𝑥𝑑𝑥 = 𝑑𝑡

∴ ∫𝑥

√𝑥2 − 1𝑑𝑥 =

1

2∫

𝑑𝑡

√𝑡

=1

2∫ 𝑡−

12𝑑𝑡

=1

2[2𝑡

12]

= √𝑡

= √𝑥2 − 1 [Substituting 𝑡] [1

Mark]

Hence, we get,

∫𝑥−1

√𝑥2−1𝑑𝑥 = ∫

𝑥

√𝑥2−1𝑑𝑥 − ∫

1


= √𝑥2 − 1 − log|𝑥 + √𝑥2 − 1| + C [∫1

√𝑥2−𝑎2𝑑𝑡 = log|𝑥 + √𝑥2 − 𝑎2|] [2

Marks]

Hence, integration of 𝑥−1

√𝑥2−1= √𝑥2 − 1 − log|𝑥 + √𝑥2 − 1| + C

[𝟏

𝟐 Mark]




√𝑥6+𝑎6 [4

Marks]

Solution:

We need to integrate the function 𝑥2

√𝑥6+𝑎6

Let 𝑥3 = 𝑡

⇒ 3𝑥2𝑑𝑥 = 𝑑𝑡 [1

Mark]

∴ ∫𝑥2

√𝑥6+𝑎6𝑑𝑥 =

1

3∫

𝑑𝑡

√𝑡2+(𝑎3)2 [

𝟏

𝟐

Mark]

=1

3log|𝑡 + √𝑡2 + 𝑎6| + 𝐶 [∫

1

√𝑥2+𝑎2𝑑𝑥 = log|𝑥 + √𝑥2 + 𝑎2|]

=1

3log|𝑥3 + √𝑥6 + 𝑎6| + 𝐶 [Substituting 𝑡] [2

Marks]

Hence, integration of 𝑥2

√𝑥6+𝑎6=

1

3log|𝑥3 + √𝑥6 + 𝑎6| + 𝐶 [

𝟏

𝟐

Mark]

9. Integrate the function sec2𝑥

√tan2𝑥+4 [4 Marks]

Solution:

We need to integrate the function sec2𝑥

√tan2𝑥+4

Let tan 𝑥 = 𝑡

∴ sec2 𝑥 𝑑𝑥 = 𝑑𝑡 [1

Mark]

⇒ ∫sec2𝑥

√tan2𝑥+4𝑑𝑥 = ∫

𝑑𝑡

√𝑡2+22 [

𝟏

𝟐

Mark]

= log|𝑡 + √𝑡2 + 4| + C [∫1

√𝑥2+𝑎2𝑑𝑥 = log|𝑥 + √𝑥2 + 𝑎2|]



= log|tan𝑥 + √tan2𝑥 + 4| + C [Substituting 𝑡] [2

Marks]

Hence, integration of sec2𝑥

√tan2𝑥+4= log|tan𝑥 + √tan2𝑥 + 4| + C

[𝟏

𝟐 Mark]


√𝑥2+2𝑥+2 [4 Marks]

Solution:


√𝑥2+2𝑥+2

∫1

√𝑥2+2𝑥+2𝑑𝑥 = ∫

1

√(𝑥+1)2+(1)2𝑑𝑥

[𝟏

𝟐 Mark]

Let 𝑥 + 1 = 𝑡

∴ 𝑑𝑥 = 𝑑𝑥 [1

Mark]

⇒ ∫1

√𝑥2 + 2𝑥 + 2𝑑𝑥 = ∫

1

√𝑡2 + 1𝑑𝑡

= log|𝑡 + √𝑡2 + 1| + C [∫1

√𝑥2+𝑎2𝑑𝑥 = log|𝑥 + √𝑥2 + 𝑎2|]

= log |(𝑥 + 1) + √(𝑥 + 1)2 + 1| + C [Substituting 𝑡]

= log|(𝑥 + 1) + √𝑥2 + 2𝑥 + 2| + C [2

Marks]


√𝑥2+2𝑥+2= log|(𝑥 + 1) + √𝑥2 + 2𝑥 + 2| + C

[𝟏

𝟐 Mark]


9𝑥2+6𝑥+5 [4 Marks]

Solution:




9𝑥2+6𝑥+5

∫1

9𝑥2+6𝑥+5𝑑𝑥 = ∫

1

(3𝑥+1)2+(2)2 𝑑𝑥 [𝟏

𝟐

Mark]

Let 3𝑥 + 1 = 𝑡

∴ 3𝑑𝑥 = 𝑑𝑡 [1

Mark]

⇒ ∫1

(3𝑥 + 1)2 + (2)2𝑑𝑥 =

1

3∫

1

𝑡2 + 22𝑑𝑡

=1

3×

1

2tan−1 𝑡

2+ 𝐶 [Since, ∫

1

𝑥2+𝑎2 𝑑𝑥 =1

𝑎tan−1 𝑥

𝑎+ 𝐶]

=1

6tan−1 (

3𝑥+1

2) + C [Substituting 𝑡] [2

Marks]


9𝑥2+6𝑥+5=

1

6tan−1 (

3𝑥+1

2) + C

[𝟏

𝟐 Mark]


√7−6𝑥−𝑥2 [4 Marks]

Solution:


√7−6𝑥−𝑥2

7 − 6𝑥 − 𝑥2 can be written as 7 − (𝑥2 + 6𝑥 + 9 − 9)

Hence,

7 − (𝑥2 + 6𝑥 + 9 − 9)

= 16 − (𝑥2 + 6𝑥 + 9)

= 16 − (𝑥 + 3)2

= (4)2 − (𝑥 + 3)2 [𝟏

𝟐

Mark]

∴ ∫1

√7 − 6𝑥 − 𝑥2𝑑𝑥 = ∫

1

√(4)2 − (𝑥 + 3)2𝑑𝑥

Let 𝑥 + 3 = 𝑡



⇒ 𝑑𝑥 = 𝑑𝑡 [1

Mark]

⇒ ∫1

√(4)2 − (𝑥 + 3)2𝑑𝑥 = ∫

1

√(4)2 − (𝑡)2𝑑𝑡

= sin−1 (𝑡

4) + 𝐶 [Since, ∫

1


𝑎+ 𝐶]

= sin−1 (𝑥+3

4) + 𝐶 [Substituting 𝑡] [2

Marks]


√7−6𝑥−𝑥2= sin−1 (

𝑥+3

4) + 𝐶

[𝟏

𝟐 Mark]


√(𝑥−1)(𝑥−2) [4

Marks]

Solution:


√(𝑥−1)(𝑥−2)

(𝑥 − 1)(𝑥 − 2) can be written as 𝑥2 − 3𝑥 + 2

Hence,

𝑥2 − 3𝑥 + 2

= 𝑥2 − 3𝑥 +9

4−

9

4+ 2

= (𝑥 −3

2)

2−

1

4

= (𝑥 −3

2)

2− (

1

2)

2 [1

Mark]

∴ ∫1

√(𝑥−1)(𝑥−2)𝑑𝑥 = ∫

1

√(𝑥−3

2)

2−(

1

2)

2𝑑𝑥

Let 𝑥 −3

2= 𝑡




⇒ ∫1

√(𝑥−3

2)

2−(

1

2)

2𝑑𝑥 = ∫

1

√𝑡2−(1

2)

2𝑑𝑡 [

𝟏

𝟐

Mark]

= log [𝑡 + √𝑡2 − (1

2)

2 ] + 𝐶 [∫

1

√𝑥2−𝑎2𝑑𝑥 = log|𝑥 + √𝑥2 − 𝑎2|]

= log |(𝑥 −3

2) + √𝑥2 − 3𝑥 + 2| + 𝐶 [Substituting 𝑡] [2

Marks]


√(𝑥−1)(𝑥−2)= log |(𝑥 −

3

2) + √𝑥2 − 3𝑥 + 2| + 𝐶

[𝟏

𝟐 Mark]


√8+3𝑥−𝑥2 [4 Marks]

Solution:


√8+3𝑥−𝑥2

8 + 3𝑥 − 𝑥2 can be written as 8 − (𝑥2 − 3𝑥 +9

4−

9

4)

Hence,

8 − (𝑥2 − 3𝑥 +9

4−

9

4)

=41

4− (𝑥 −

3

2)

2 [1

Mark]

⇒ ∫1

√8 + 3𝑥 − 𝑥2𝑑𝑥 = ∫

1

√414 − (𝑥 −

32)

2𝑑𝑥

Let 𝑥 −3

2= 𝑡


⇒ ∫1

√41

4−(𝑥−

3

2)

2𝑑𝑥 = ∫

1

√(√41

2)

2

−𝑡2

𝑑𝑡 [1

Mark]



= sin−1 (t

√41

2

) + C [Since, ∫1


𝑎]

= sin−1 (2𝑥−3

√41) + C [Substituting 𝑡]


√8+3𝑥−𝑥2= sin−1 (

2𝑥−3

√41) +C [2

Marks]


√(𝑥−𝑎)(𝑥−𝑏) [4

Marks]

Solution:


√(𝑥−𝑎)(𝑥−𝑏)

(𝑥 − 𝑎)(𝑥 − 𝑏) can be written as 𝑥2 − (𝑎 + 𝑏)𝑥 + 𝑎𝑏

Therefore,

𝑥2 − (𝑎 + 𝑏)𝑥 + 𝑎𝑏

= 𝑥2 − (𝑎 + 𝑏)𝑥 +(𝑎 + 𝑏)2

4−

(𝑎 + 𝑏)2

4+ 𝑎𝑏

= [𝑥 − (𝑎+𝑏

2)]

2−

(𝑎−𝑏)2

4 [1

Mark]

⇒ ∫1

√(𝑥 − 𝑎)(𝑥 − 𝑏)𝑑𝑥 = ∫

1

√{𝑥 − (𝑎 + 𝑏

2 )}2

− (𝑎 − 𝑏

2 )2

𝑑𝑥

Let 𝑥 − (𝑎+𝑏

2) = 𝑡


⇒ ∫1

√{𝑥−(𝑎+𝑏

2)}

2−(

𝑎−𝑏

2)

2𝑑𝑥 = ∫

1

√𝑡2−(𝑎−𝑏

2)

2𝑑𝑡 [1

Mark]

= log |𝑡 + √𝑡2 − (𝑎−𝑏

2)

2| + C [∫

1

√𝑥2−𝑎2𝑑𝑥 = log|𝑥 + √𝑥2 − 𝑎2|]

= log |𝑥 − (𝑎+𝑏

2) + √(𝑥 − 𝑎)(𝑥 − 𝑏)| + C [Substituting 𝑡]




√(𝑥−𝑎)(𝑥−𝑏)= log |𝑥 − (

𝑎+𝑏

2) + √(𝑥 − 𝑎)(𝑥 − 𝑏)| + C [2

Marks]

16. Integrate the function 4𝑥+1

√2𝑥2+𝑥−3 [4 Marks]

Solution:

We need to integrate the function 4𝑥+1

√2𝑥2+𝑥−3

Let 4𝑥 + 1 = 𝐴𝑑

𝑑𝑥(2𝑥2 + 𝑥 − 3) + 𝐵

⇒ 4𝑥 + 1 = 𝐴(4𝑥 + 1) + 𝐵

⇒ 4𝑥 + 1 = 4𝐴𝑥 + 𝐴 + 𝐵

Equating the coefficients of 𝑥 and constant on both sides, we get

4𝐴 = 4 ⇒ 𝐴 = 1

𝐴 + 𝐵 = 1 ⇒ 𝐵 = 0 [2

Marks]

Let 2𝑥2 + 𝑥 − 3 = 𝑡

∴ (4𝑥 + 1)𝑑𝑥 = 𝑑𝑡

⇒ ∫4𝑥 + 1

√2𝑥2 + 𝑥 − 3𝑑𝑥 = ∫

1

√𝑡𝑑𝑡

= 2√𝑡 + C

= 2√2𝑥2 + 𝑥 − 3 + C [Substituting 𝑡]

Hence, integration of 4𝑥+1

√2𝑥2+𝑥−3= 2√2𝑥2 + 𝑥 − 3 + C [2

Marks]

17. Integrate the function 𝑥+2

√𝑥2−1 [4

Marks]

Solution:



We need to integrate the function 𝑥+2

√𝑥2−1

Let 𝑥 + 2 = 𝐴𝑑

𝑑𝑥(𝑥2 − 1) + 𝐵 … (1)

⇒ 𝑥 + 2 = 𝐴(2𝑥) + 𝐵

Equating the coefficients of 𝑥 and constant term on both sides, we obtain

2𝐴 = 1 ⇒ 𝐴 =1

2

𝐵 = 2

From (1), we obtain

(𝑥 + 2) =1

2(2𝑥) + 2 [1

Mark]

Then, ∫𝑥+2

√𝑥2−1𝑑𝑥 = ∫

1

2(2𝑥)+2


=1

2∫

2𝑥

√𝑥2 − 1𝑑𝑥 + ∫

2

√𝑥2 − 1𝑑𝑥

For 1

2∫

2𝑥

√𝑥2−1𝑑𝑥, Let 𝑥2 − 1 = 𝑡 ⇒ 2𝑥𝑑𝑥 = 𝑑𝑡

Hence, 1

2∫

2𝑥

√𝑥2−1𝑑𝑥 =

1

2∫

𝑑𝑡

√𝑡

=1

2[2√𝑡]

= √𝑡

= √𝑥2 − 1…….(2) [Substituting 𝑡] [1

Mark]

Again, ∫2

√𝑥2−1𝑑𝑥 = 2 ∫

1

√𝑥2−1𝑑𝑥 = 2 log|𝑥 + √𝑥2 − 1| [1

Mark]

Hence, ∫𝑥+2

√𝑥2−1𝑑𝑥 = √𝑥2 − 1 + 2log|𝑥 + √𝑥2 − 1| + C [1

Mark]

18. Integrate the function 5𝑥−2

1+2𝑥+3𝑥2 [6 Marks]

Solution:



We need to integrate the function 5𝑥−2

1+2𝑥+3𝑥2

Let 5𝑥 − 2 = 𝐴𝑑

𝑑𝑥(1 + 2𝑥 + 3𝑥2) + 𝐵

⇒ 5𝑥 − 2 = 𝐴(2 + 6𝑥) + 𝐵

Equating the coefficient of 𝑥 and constant term on both sides, we obtain

5 = 6𝐴 ⇒ 𝐴 =5

6

2𝐴 + 𝐵 = −2 ⇒ 𝐵 = −11

3

∴ 5𝑥 − 2 =5

6(2 + 6𝑥) + (−

11

3) [1

Mark]

⇒ ∫5𝑥 − 2

1 + 2𝑥 + 3𝑥2𝑑𝑥 = ∫

56

(2 + 6𝑥) −113

1 + 2𝑥 + 3𝑥2𝑑𝑥

=5

6∫

2 + 6𝑥

1 + 2𝑥 + 3𝑥2𝑑𝑥 −

11

3∫

1

1 + 2𝑥 + 3𝑥2𝑑𝑥

Let 𝐼1 = ∫2+6𝑥

1+2𝑥+3𝑥2 𝑑𝑥 and 𝐼2 = ∫1

1+2𝑥+3𝑥2 𝑑𝑥

∴ ∫5𝑥−2

1+2𝑥+3𝑥2 𝑑𝑥 =5

6𝐼1 −

11

3𝐼2 … (1) [1

Mark]

𝐼1 = ∫2 + 6𝑥

1 + 2𝑥 + 3𝑥2𝑑𝑥

Let 1 + 2𝑥 + 3𝑥2 = 𝑡 [𝟏

𝟐

Mark]

⇒ (2 + 6𝑥)𝑑𝑥 = 𝑑𝑡

∴ 𝐼1 = ∫𝑑𝑡

𝑡

𝐼1 = log|𝑡|

𝐼1 = log|1 + 2𝑥 + 3𝑥2| … (2) [Substituting 𝑡] [1

Mark]

For 𝐼2 = ∫1

1+2𝑥+3𝑥2 𝑑𝑥

1 + 2𝑥 + 3𝑥2 = 1 + 3 (𝑥2 +2

3𝑥)

= 1 + 3 (𝑥2 +2

3𝑥 +

1

9−

1

9)



= 1 + 3 (𝑥 +1

3)

2

−1

3

=2

3+ 3 (𝑥 +

1

3)

2

= 3 [(𝑥 +1

3)

2

+2

9]

= 3 [(𝑥 +1

3)

2

+ (√2

3)

2

]

Hence, 𝐼2 =1

3∫

1

[(𝑥+1

3)

2+(

√2

3)

2

]

𝑑𝑥

=1

3[

1

√2

3

tan−1 (𝑥+

1

3

√2

3

)] [Since, ∫1

𝑥2+𝑎2 𝑑𝑥 =1

𝑎tan−1 𝑥

𝑎+ 𝐶]

=1

3[

3

√2tan−1 (

3𝑥 + 1

√2)]

=1

√2tan−1 (

3𝑥+1

√2) … (3) [2

Marks]

Substituting equations (2) and (3) in equation (1), we obtain

∫5𝑥 − 2

1 + 2𝑥 + 3𝑥2𝑑𝑥 =

5

6[log|1 + 2𝑥 + 3𝑥2|] −

11

3[

1

√2tan−1 (

3𝑥 + 1

√2)] + C

=5

6log|1 + 2𝑥 + 3𝑥2| −

11

3√2tan−1 (

3𝑥+1

√2) + C

Hence, integration of 5𝑥−2

1+2𝑥+3𝑥2 =5

6log|1 + 2𝑥 + 3𝑥2| −

11

3√2tan−1 (

3𝑥+1

√2) + C

[𝟏

𝟐 Mark]


√(𝑥−5)(𝑥−4) [6

Marks]

Solution:


√(𝑥−5)(𝑥−4)

6𝑥+7

√(𝑥−5)(𝑥−4)=

6𝑥+7

√𝑥2−9𝑥+20




𝑑𝑥(𝑥2 − 9𝑥 + 20) + 𝐵

⇒ 6𝑥 + 7 = 𝐴(2𝑥 − 9) + 𝐵

Equating the coefficients of 𝑥 and constant term, we obtain

2𝐴 = 6 ⇒ 𝐴 = 3

−9𝐴 + 𝐵 = 7 ⇒ 𝐵 = 34

∴ 6𝑥 + 7 = 3(2𝑥 − 9) + 34 [1

Mark]

∫6𝑥 + 7

√𝑥2 − 9𝑥 + 20= ∫

3(2𝑥 − 9) + 34

√𝑥2 − 9𝑥 + 20𝑑𝑥

= 3 ∫2𝑥−9

√𝑥2−9𝑥+20𝑑𝑥 + 34 ∫

1

√𝑥2−9𝑥+20𝑑𝑥

Let 𝐼1 = ∫2𝑥−9

√𝑥2−9𝑥+20 and 𝐼2 = ∫

1

√𝑥2−9𝑥+20𝑑𝑥

∴ ∫6𝑥+7

√𝑥2−9𝑥+20= 3𝐼1 + 34𝐼2 … (1) [1

Mark]

Now,

𝐼1 = ∫2𝑥 − 9

√𝑥2 − 9𝑥 + 20𝑑𝑥

Let 𝑥2 − 9𝑥 + 20 = 𝑡 [𝟏

𝟐

Mark]

⇒ (2𝑥 − 9)𝑑𝑥 = 𝑑𝑡

⇒ 𝐼1 = ∫𝑑𝑡

√𝑡

= 2√𝑡

Hence, 𝐼1 = 2√𝑥2 − 9𝑥 + 20 … (2) [Substituting 𝑡] [1

Mark]

and 𝐼2 = ∫1

√𝑥2−9𝑥+20𝑑𝑥

𝑥2 − 9𝑥 + 20 can be written as 𝑥2 − 9𝑥 + 20 +81

4−

81

4.

Hence,

𝑥2 − 9𝑥 + 20 +81

4−

81

4



= (𝑥 −9

2)

2

−1

4

= (𝑥 −9

2)

2

− (1

2)

2

⇒ 𝐼2 = ∫1

√(𝑥 −92)

2

− (12)

2𝑑𝑥

= log |(𝑥 −9

2) + √𝑥2 − 9𝑥 + 20| … (3) [∫

1

√𝑥2−𝑎2𝑑𝑥 = log|𝑥 + √𝑥2 − 𝑎2|] [2

Marks]

Substituting equations (2) and (3) in (1), we obtain

∫6𝑥 + 7

√𝑥2 − 9𝑥 + 20𝑑𝑥 = 3 [2√𝑥2 − 9𝑥 + 20] + 34 log [(𝑥 −

9

2) + √𝑥2 − 9𝑥 + 20] + C

= 6√𝑥2 − 9𝑥 + 20 + 34 log |(𝑥 −9

2) + √𝑥2 − 9𝑥 + 20| + C


√(𝑥−5)(𝑥−4)= 6√𝑥2 − 9𝑥 + 20 + 34 log |(𝑥 −

9

2) + √𝑥2 − 9𝑥 + 20| + C

[𝟏

𝟐 Mark]


√4𝑥−𝑥2 [6 Marks]

Solution:


√4𝑥−𝑥2

Let 𝑥 + 2 = 𝐴𝑑

𝑑𝑥(4𝑥 − 𝑥2) + 𝐵

⇒ 𝑥 + 2 = 𝐴(4 − 2𝑥) + 𝐵


−2𝐴 = 1 ⇒ 𝐴 = −1

2

4𝐴 + 𝐵 = 2 ⇒ 𝐵 = 4

⇒ (𝑥 + 2) = −1

2(4 − 2𝑥) + 4 [1

Mark]

∴ ∫𝑥+2

√4𝑥−𝑥2𝑑𝑥 = ∫

−1

2(4−2𝑥)+4

√4𝑥−𝑥2𝑑𝑥



= −1

2∫

4−2𝑥

√4𝑥−𝑥2𝑑𝑥 + 4 ∫

1


Let 𝐼1 = ∫4−2𝑥

√4𝑥−𝑥2𝑑𝑥 and 𝐼2 = ∫

1

√4𝑥−𝑥2𝑑𝑥 … (1)

∴ ∫𝑥+2

√4𝑥−𝑥2𝑑𝑥 = −

1

2𝐼1 + 4𝐼2 [1

Mark]

Now, 𝐼1 = ∫4−2𝑥


Let 4𝑥 − 𝑥2 = 𝑡

⇒ (4 − 2𝑥)𝑑𝑥 = 𝑑𝑡

⇒ 𝐼1 = ∫1

√𝑡𝑑𝑡 = 2√𝑡 = 2√4𝑥 − 𝑥2 …(2) [Substituting 𝑡] [1

Mark]

For 𝐼2 = ∫1


4𝑥 − 𝑥2 = −(−4𝑥 + 𝑥2)

= −(−4𝑥 + 𝑥2 + 4 − 4)

= 4 − (𝑥 − 2)2

= (2)2 − (𝑥 − 2)2 [𝟏

𝟐

Mark]

Hence, 𝐼2 = ∫1

√(2)2−(𝑥−2)2𝑑𝑥

= sin−1 (𝑥−2

2) …(3) [Since, ∫

1


𝑎+ 𝐶] [1

Mark]

Using equations (2) and (3) in (1), we get

∫𝑥 + 2

√4𝑥 − 𝑥2𝑑𝑥 = −

1

2(2√4𝑥 − 𝑥2) + 4sin−1 (

𝑥 − 2

2) + C

= −√4𝑥 − 𝑥2 + 4sin−1 (𝑥−2

2) + C [1

Mark]

Hence, integration of 𝑥+2

√4𝑥−𝑥2= −√4𝑥 − 𝑥2 + 4sin−1 (

𝑥−2

2) + C

[𝟏

𝟐 Mark]


√𝑥2+2𝑥+3 [6 Marks]



Solution:


√𝑥2+2𝑥+3

∫(𝑥+2)

√𝑥2+2𝑥+3𝑑𝑥 =

1

2∫

2(𝑥+2)

√𝑥2+2𝑥+3𝑑𝑥

=1

2∫

2𝑥 + 4

√𝑥2 + 2𝑥 + 3𝑑𝑥

=1

2∫

2𝑥 + 2

√𝑥2 + 2𝑥 + 3𝑑𝑥 +

1

2∫

2

√𝑥2 + 2𝑥 + 3𝑑𝑥

=1

2∫

2𝑥+2

√𝑥2+2𝑥+3𝑑𝑥 + ∫

1

√𝑥2+2𝑥+3𝑑𝑥 [1

Mark]

Let 𝐼1 = ∫2𝑥+2

√𝑥2+2𝑥+3𝑑𝑥 and 𝐼2 = ∫

1


∴ ∫𝑥+2

√𝑥2+2𝑥+3𝑑𝑥 =

1

2𝐼1 + 𝐼2 … (1) [1

Mark]

Now, 𝐼1 = ∫2𝑥+2


Let 𝑥2 + 2𝑥 + 3 = 𝑡

⇒ (2𝑥 + 2)𝑑𝑥 = 𝑑𝑡

𝐼1 = ∫𝑑𝑡

√𝑡= 2√𝑡 = 2√𝑥2 + 2𝑥 + 3 … (2) [Substituting 𝑡] [1

Mark]

𝐼2 = ∫1


⇒ 𝑥2 + 2𝑥 + 3 = 𝑥2 + 2𝑥 + 1 + 2 = (𝑥 + 1)2 + (√2)2

[𝟏

𝟐

Mark]

∴ 𝐼2 = ∫1

√(𝑥+1)2+(√2)2

𝑑𝑥 = log|(𝑥 + 1) + √𝑥2 + 2𝑥 + 3| … (3) [1

Mark]

Using equations (2) and (3) in (1), we obtain

∫𝑥 + 2

√𝑥2 + 2𝑥 + 3𝑑𝑥 =

1

2[2√𝑥2 + 2𝑥 + 3] + log |(𝑥 + 1) + √𝑥2 + 2𝑥 + 3| + C

= √𝑥2 + 2𝑥 + 3 + log|(𝑥 + 1) + √𝑥2 + 2𝑥 + 3| + C [1

Mark]




√𝑥2+2𝑥+3= √𝑥2 + 2𝑥 + 3 + log|(𝑥 + 1) + √𝑥2 + 2𝑥 + 3| + C

[𝟏

𝟐 Mark]


𝑥2−2𝑥−5 [6 Marks]

Solution:


𝑥2−2𝑥−5

Let (𝑥 + 3) = 𝐴𝑑

𝑑𝑥(𝑥2 − 2𝑥 − 5) + 𝐵

(𝑥 + 3) = 𝐴(2𝑥 − 2) + 𝐵


2𝐴 = 1 ⇒ 𝐴 =1

2

−2𝐴 + 𝐵 = 3 ⇒ 𝐵 = 4

∴ (𝑥 + 3) =1

2(2𝑥 − 2) + 4 [1

Mark]

⇒ ∫𝑥 + 3

𝑥2 − 2𝑥 − 5𝑑𝑥 = ∫

12

(2𝑥 − 2) + 4

𝑥2 − 2𝑥 − 5𝑑𝑥

=1

2∫

2𝑥 − 2

𝑥2 − 2𝑥 − 5𝑑𝑥 + 4 ∫

1

𝑥2 − 2𝑥 − 5𝑑𝑥

Let 𝐼1 = ∫2𝑥−2

𝑥2−2𝑥−5𝑑𝑥 and 𝐼2 = ∫

1

𝑥2−2𝑥−5𝑑𝑥

∴ ∫ 𝑥+3

(𝑥2−2𝑥−5)𝑑𝑥 =

1

2𝐼1 + 4𝐼2 … (1) [1

Mark]

Now, 𝐼1 = ∫2𝑥−2

𝑥2−2𝑥−5𝑑𝑥

Let 𝑥2 − 2𝑥 − 5 = 𝑡

⇒ (2𝑥 − 2)𝑑𝑥 = 𝑑𝑡 [𝟏

𝟐

Mark]

⇒ 𝐼1 = ∫𝑑𝑡

𝑡= log|𝑡| = log|𝑥2 − 2𝑥 − 5| … (2) [Substituting 𝑡] [1

Mark]



𝐼2 = ∫1

𝑥2 − 2𝑥 − 5𝑑𝑥

= ∫1

(𝑥2 − 2𝑥 + 1) − 6𝑑𝑥

= ∫1

(𝑥 − 1)2 − (√6)2 𝑑𝑥

=1

2√6log (

𝑥−1−√6

𝑥−1+√6) … (3) [1

Mark]

Substituting (2) and (3) in (1), we obtain

∫𝑥 + 3

𝑥2 − 2𝑥 − 5𝑑𝑥 =

1

2log|𝑥2 − 2𝑥 − 5| +

4

2√6log |

𝑥 − 1 − √6

𝑥 − 1 + √6| + 𝐶

=1

2log|𝑥2 − 2𝑥 − 5| +

2

√6log |

𝑥−1−√6

𝑥−1+√6| + 𝐶 [1

Mark]


𝑥2−2𝑥−5=

1

2log|𝑥2 − 2𝑥 − 5| +

2

√6log |

𝑥−1−√6

𝑥−1+√6| + 𝐶

[𝟏

𝟐 Mark]


√𝑥2+4𝑥+10 [6

Marks]

Solution:


√𝑥2+4𝑥+10


𝑑𝑥(𝑥2 + 4𝑥 + 10) + 𝐵

⇒ 5𝑥 + 3 = 𝐴(2𝑥 + 4) + 𝐵

Equating the coefficients of 𝑥 and constant term, we obtain

2𝐴 = 5 ⇒ 𝐴 =5

2

4𝐴 + 𝐵 = 3 ⇒ 𝐵 = −7

∴ 5𝑥 + 3 =5

2(2𝑥 + 4) − 7 [1

Mark]



⇒ ∫5𝑥 + 3

√𝑥2 + 4𝑥 + 10𝑑𝑥 = ∫

52

(2𝑥 + 4) − 7

√𝑥2 + 4𝑥 + 10𝑑𝑥

=5

2∫

2𝑥 + 4

√𝑥2 + 4𝑥 + 10𝑑𝑥 − 7 ∫

1

√𝑥2 + 4𝑥 + 10𝑑𝑥

Let 𝐼1 = ∫2𝑥+4

√𝑥2+4𝑥+10𝑑𝑥 and 𝐼2 = ∫

1

√𝑥2+4𝑥+10𝑑𝑥

∴ ∫5𝑥+3

√𝑥2+4𝑥+10𝑑𝑥 =

5

2𝐼1 − 7𝐼2 … (1) [1

Mark]

Now, 𝐼1 = ∫2𝑥+4

√𝑥2+4𝑥+10𝑑𝑥

Let 𝑥2 + 4𝑥 + 10 = 𝑡

∴ (2𝑥 + 4)𝑑𝑥 = 𝑑𝑡 [𝟏

𝟐

Mark]

⇒ 𝐼1 = ∫𝑑𝑡

𝑡= 2√𝑡 = 2√𝑥2 + 4𝑥 + 10 … (2) [Substituting 𝑡] [1

Mark]

𝐼2 = ∫1

√𝑥2 + 4𝑥 + 10𝑑𝑥

= ∫1

√(𝑥2 + 4𝑥 + 4) + 6𝑑𝑥

= ∫1

(𝑥 + 2)2 + (√6)2 𝑑𝑥

= log|(𝑥 + 2)√𝑥2 + 4𝑥 + 10| … (3) [1

Mark]

Using equations (2) and (3) in (1), we obtain

∫5𝑥 + 3

√𝑥2 + 4𝑥 + 10𝑑𝑥 =

5

2[2√𝑥2 + 4𝑥 + 10] − 7 log |(𝑥 + 2) + √𝑥2 + 4𝑥 + 10| + C

= 5√𝑥2 + 4𝑥 + 10 − 7 log|(𝑥 + 2) + √𝑥2 + 4𝑥 + 10| + C [1

Mark]


√𝑥2+4𝑥+10= 5√𝑥2 + 4𝑥 + 10 − 7 log|(𝑥 + 2) + √𝑥2 + 4𝑥 + 10| + C

[𝟏

𝟐 Mark]



24. ∫𝑑𝑥

𝑥2+2𝑥+2 equals [2

Marks]

(A) 𝑥 tan−1(𝑥 + 1) + 𝐶

(B) tan−1(𝑥 + 1) + 𝐶

(C) (𝑥 + 1) tan−1 𝑥 + 𝐶

(D) tan−1 𝑥 + 𝐶

Solution:

Given: ∫𝑑𝑥

𝑥2+2𝑥+2= ∫

𝑑𝑥

(𝑥2+2𝑥+1)+1

= ∫1

(𝑥+1)2+(1)2 𝑑𝑥

[𝟏

𝟐 Mark]

Hence, ∫𝑑𝑥

𝑥2+2𝑥+2 = [tan−1(𝑥 + 1)] + C [1

Mark]

Hence, the correct answer is 𝐵

[𝟏

𝟐 Mark]

25. ∫𝑑𝑥

√9𝑥−4𝑥2 equals [4

Marks]

(A) 1

9sin−1 (

9𝑥−8

8) + 𝐶

(B) 1

2sin−1 (

8𝑥−9

9) + 𝐶

(C) 1

3sin−1 (

9𝑥−8

8) + 𝐶

(D) 1

2sin−1 (

9𝑥−8

9) + 𝐶

Solution:

Given: ∫𝑑𝑥

√9𝑥−4𝑥2



= ∫1

√−4(𝑥2−9

4𝑥)

𝑑𝑥

= ∫1

−4(𝑥2−9

4𝑥+

81

64−

81

64)

𝑑𝑥

= ∫1

√−4[(𝑥−9

8)

2−(

9

8)

2]

𝑑𝑥

=1

2∫

1

√(9

8)

2−(𝑥−

9

8)

2𝑑𝑥 [1

Mark]

=1

2[sin−1 (

𝑥−9

89

8

)] + C (∫𝑑𝑦

√𝑎2−𝑦2= sin−1 𝑦

𝑎+ 𝐶)

=1

2sin−1 (

8𝑥−9

9) + C [2

Marks]

Hence, ∫𝑑𝑥

√9𝑥−4𝑥2=

1

2sin−1 (

8𝑥−9

9) + C

Hence, the correct answer is B. [1

Mark]

Exercise 7.5

1. Integrate the rational function 𝑥

(𝑥+1)(𝑥+2) [4

Marks]

Solution:


(𝑥+1)(𝑥+2)

Let 𝑥

(𝑥+1)(𝑥+2)=

𝐴

(𝑥+1)+

𝐵

𝑥+2 [

𝟏

𝟐

Mark]

⇒ 𝑥 = 𝐴(𝑥 + 2) + 𝐵(𝑥 + 1)

Equating the coefficients of 𝑥 and constant term, we get

𝐴 + 𝐵 = 1



2𝐴 + 𝐵 = 0

On solving we get

𝐴 = −1 and 𝐵 = 2 [1

Mark]

∴𝑥

(𝑥+1)(𝑥+2)=

−1

(𝑥+1)+

2

(𝑥+2) [

𝟏

𝟐

Mark]

⇒ ∫𝑥

(𝑥 + 1)(𝑥 + 2)𝑑𝑥 = ∫

−1

(𝑥 + 1)+

2

(𝑥 + 2)𝑑𝑥

= − log|𝑥 + 1| + 2 log|𝑥 + 2| + C

= log(𝑥 + 2)2 − log|𝑥 + 1| + C

= log(𝑥+2)2

|𝑥+1|+ C

Hence, Integration of 𝑥

(𝑥+1)(𝑥+2)= log

(𝑥+2)2

|𝑥+1|+ C [2

Marks]

2. Integrate the rational function 1

𝑥2−9 [4

Marks]

Solution:


𝑥2−9

Let 1

(𝑥+3)(𝑥−3)=

𝐴

(𝑥+3)+

𝐵

(𝑥−3) [

𝟏

𝟐

Mark]

⇒ 1 = 𝐴(𝑥 − 3) + 𝐵(𝑥 + 3)


𝐴 + 𝐵 = 0

−3𝐴 + 3𝐵 = 1

On solving, we get

𝐴 = −1

6 and 𝐵 =

1

6 [1

Mark]



∴1

(𝑥+3)(𝑥−3)=

−1

6(𝑥+3)+

1

6(𝑥−3) [

𝟏

𝟐

Mark]

⇒ ∫1

(𝑥2 − 9)𝑑𝑥 = ∫ (

−1

6(𝑥 + 3)+

1

6(𝑥 − 3)) 𝑑𝑥

= −1

6log|𝑥 + 3| +

1

6log|𝑥 − 3| + C

=1

6log |

(𝑥−3)

(𝑥+3)| + C

Hence, Integration of 1

𝑥2−9=

1

6log |

(𝑥−3)

(𝑥+3)| + C [2

Marks]

3. Integrate the rational function 3𝑥−1

(𝑥−1)(𝑥−2)(𝑥−3) [4

Marks]

Solution:

We need to integrate 3𝑥−1

(𝑥−1)(𝑥−2)(𝑥−3)

Let 3𝑥−1

(𝑥−1)(𝑥−2)(𝑥−3)=

𝐴

(𝑥−1)+

𝐵

(𝑥−2)+

𝐶

(𝑥−3) [

𝟏

𝟐

Mark]

3𝑥 − 1 = 𝐴(𝑥 − 2)(𝑥 − 3) + 𝐵(𝑥 − 1)(𝑥 − 3) + 𝐶(𝑥 − 1)(𝑥 − 2) … (1)

Equating the coefficients of 𝑥2, 𝑥 and constant term, we get

𝐴 + 𝐵 + 𝐶 = 0

– 5𝐴– 4𝐵 – 3𝐶 = 3

6𝐴 + 3𝐵 + 2𝐶 =– 1

Solving these equations, we get

𝐴 = 1, 𝐵 = −5, and 𝐶 = 4 [1

Mark]

∴3𝑥−1

(𝑥−1)(𝑥−2)(𝑥−3)=

1

(𝑥−1)−

5

(𝑥−2)+

4

(𝑥−3) [

𝟏

𝟐

Mark]

⇒ ∫3𝑥 − 1

(𝑥 − 1)(𝑥 − 2)(𝑥 − 3)𝑑𝑥 = ∫ {

1

(𝑥 − 1)−

5

(𝑥 − 2)+

4

(𝑥 − 3)} 𝑑𝑥



= log|𝑥 − 1| − 5 log|𝑥 − 2| + 4 log|𝑥 − 3| + 𝐶

Hence, Integration of 3𝑥−1

(𝑥−1)(𝑥−2)(𝑥−3)= log|𝑥 − 1| − 5 log|𝑥 − 2| + 4 log|𝑥 − 3| + 𝐶 [2

Marks]


(𝑥−1)(𝑥−2)(𝑥−3) [4

Marks]

Solution:


(𝑥−1)(𝑥−2)(𝑥−3)

Let 𝑥

(𝑥−1)(𝑥−2)(𝑥−3)=

𝐴

(𝑥−1)+

𝐵

(𝑥−2)+

𝐶

(𝑥−3) [

𝟏

𝟐

Mark]

⇒ 𝑥 = 𝐴(𝑥 − 2)(𝑥 − 3) + 𝐵(𝑥 − 1)(𝑥 − 3) + 𝐶(𝑥 − 1)(𝑥 − 2) … (1)


𝐴 + 𝐵 + 𝐶 = 0

– 5𝐴– 4𝐵– 3𝐶 = 1

6𝐴 + 4𝐵 + 2𝐶 = 0


𝐴 =1

2, 𝐵 = −2, and 𝐶 =

3

2 [1

Mark]

∴𝑥

(𝑥−1)(𝑥−2)(𝑥−3)=

1

2(𝑥−1)−

2

(𝑥−2)+

3

2(𝑥−3) [

𝟏

𝟐

Mark]

⇒ ∫𝑥

(𝑥 − 1)(𝑥 − 2)(𝑥 − 3)𝑑𝑥 = ∫ {

1

2(𝑥 − 1)−

2

(𝑥 − 2)+

3

2(𝑥 − 3)} 𝑑𝑥

=1

2log|𝑥 − 1| − 2 log|𝑥 − 2| +

3

2log|𝑥 − 3| + C

Hence, integration of 𝑥

(𝑥−1)(𝑥−2)(𝑥−3)=

1

2log|𝑥 − 1| − 2 log|𝑥 − 2| +

3

2log|𝑥 − 3| + C [2

Marks]

5. Integrate the rational function 2𝑥

𝑥2+3𝑥+2 [4 Marks]



Solution:

We need to integrate: 2𝑥

𝑥2+3𝑥+2

Let 2𝑥

𝑥2+3𝑥+2=

𝐴

(𝑥+1)+

𝐵

(𝑥+2) [

𝟏

𝟐

Mark]

⇒ 2𝑥 = 𝐴(𝑥 + 2) + 𝐵(𝑥 + 1) … (1)


𝐴 + 𝐵 = 2

2𝐴 + 𝐵 = 0


𝐴 = −2 and 𝐵 = 4 [1

Mark]

∴2𝑥

(𝑥+1)(𝑥+2)=

−2

(𝑥+1)+

4

(𝑥+2) [

𝟏

𝟐

Mark]

⇒ ∫2𝑥

(𝑥 + 1)(𝑥 + 2)𝑑𝑥 = ∫ {

4

(𝑥 + 2)−

2

(𝑥 + 1)} 𝑑𝑥

= 4 log|𝑥 + 2| − 2 log|𝑥 + 1| + C

Hence, Integration of 2𝑥

𝑥2+3𝑥+2= 4 log|𝑥 + 2| − 2 log|𝑥 + 1| + C [2

Marks]

6. Integrate the rational function 1−𝑥2

𝑥(1−2𝑥) [4 Marks]

Solution:

We need to integrate 1−𝑥2

𝑥(1−2𝑥)

It is seen that the given integrand is not a proper fraction. Therefore, on dividing (1 − 𝑥2) by

𝑥(1 − 2𝑥), we get

1−𝑥2

𝑥(1−2𝑥)=

1

2+

1

2(

2−𝑥

𝑥(1−2𝑥)) …(1)



Let 2−𝑥

𝑥(1−2𝑥)=

𝐴

𝑥+

𝐵

(1−2𝑥) [

𝟏

𝟐

Mark]

⇒ (2 − 𝑥) = 𝐴(1 − 2𝑥) + 𝐵𝑥


– 2𝐴 + 𝐵 =– 1

And 𝐴 = 2


𝐴 = 2 and 𝐵 = 3 [1

Mark]

∴2−𝑥

𝑥(1−2𝑥)=

2

𝑥+

3

1−2𝑥 [

𝟏

𝟐

Mark]

Substituting in equation (1), we get

1 − 𝑥2

𝑥(1 − 2𝑥)=

1

2+

1

2{2

𝑥+

2

(1 − 2𝑥)}

⇒ ∫1 − 𝑥2

𝑥(1 − 2𝑥)𝑑𝑥 = ∫ {

1

2+

1

2(

2

𝑥+

3

1 − 2𝑥)} 𝑑𝑥

=𝑥

2+ log|𝑥| +

3

2(−2)log|1 − 2𝑥| + C

=𝑥

2+ log|𝑥| −

3

4log|1 − 2𝑥| + C

Hence, Integration of 1−𝑥2

𝑥(1−2𝑥)=

𝑥

2+ log|𝑥| −

3

4log|1 − 2𝑥| + C [2

Marks]


(𝑥2+1)(𝑥−1) [6

Marks]

Solution:


(𝑥2+1)(𝑥−1)



Let 𝑥

(𝑥2+1)(𝑥−1)=

𝐴𝑥+𝐵

(𝑥2+1)+

𝐶

(𝑥−1) …(1) [

𝟏

𝟐

Mark]

⇒ 𝑥 = (𝐴𝑥 + 𝐵)(𝑥 − 1) + 𝐶(𝑥2 + 1)

⇒ 𝑥 = 𝐴𝑥2 − 𝐴𝑥 + 𝐵𝑥 − 𝐵 + 𝐶𝑥2 + 𝐶

Equating the coefficient of 𝑥2, 𝑥, and constant term, we get

𝐴 + 𝐶 = 0

−𝐴 + 𝐵 = 1

−𝐵 + 𝐶 = 0

On solving these equations, we get

𝐴 = −1

2, 𝐵 =

1

2, and 𝐶 =

1

2 [2

Marks]

From equation (1), we get

∴𝑥

(𝑥2+1)(𝑥−1)=

(−1

2𝑥+

1

2)

𝑥2+1+

1

2

(𝑥−1) [

𝟏

𝟐

Mark]

⇒ ∫𝑥

(𝑥2 + 1)(𝑥 − 1)𝑑𝑥 = −

1

2∫

𝑥

𝑥2 + 1𝑑𝑥 +

1

2∫

1

𝑥2 + 1𝑑𝑥 +

1

2∫

1

𝑥 − 1𝑑𝑥

= −1

4∫

2𝑥

𝑥2 + 1𝑑𝑥 +

1

2tan−1𝑥 +

1

2log|𝑥 − 1| + C

Consider ∫2𝑥

𝑥2+1𝑑𝑥, let (𝑥2 + 1) = 𝑡 ⇒ 2𝑥𝑑𝑥 = 𝑑𝑡

⇒ ∫2𝑥

𝑥2 + 1𝑑𝑥 = ∫

𝑑𝑡

𝑡= log|𝑡| = log|𝑥2 + 1|

∴ ∫𝑥

(𝑥2 + 1)(𝑥 − 1)= −

1

4log|𝑥2 + 1| +

1

2tan−1𝑥 +

1

2log|𝑥 − 1| + C

=1

2log|𝑥 − 1| −

1

4log|𝑥2 + 1| +

1

2tan−1𝑥 + C


(𝑥2+1)(𝑥−1)=

1

2log|𝑥 − 1| −

1

4log|𝑥2 + 1| +

1

2tan−1𝑥 + C [3

Marks]


(𝑥−1)2(𝑥+2) [6

Marks]



Solution:


(𝑥−1)2(𝑥+2)

Let 𝑥

(𝑥−1)2(𝑥+2)=

𝐴

(𝑥−1)+

𝐵

(𝑥−1)2 +𝐶

(𝑥+2) [

𝟏

𝟐

Mark]

⇒ 𝑥 = 𝐴(𝑥 − 1)(𝑥 + 2) + 𝐵(𝑥 + 2) + 𝐶(𝑥 − 1)2


𝐴 + 𝐶 = 0

𝐴 + 𝐵 – 2𝐶 = 1

−2𝐴 + 2𝐵 + 𝐶 = 0

On solving, we get

𝐴 =2

9 and 𝐶 =

−2

9

𝐵 =1

3 [2

Marks]

∴𝑥

(𝑥−1)2(𝑥+2)=

2

9(𝑥−1)+

1

3(𝑥−1)2 −2

9(𝑥+2) [

𝟏

𝟐

Mark]

⇒ ∫𝑥

(𝑥 − 1)2(𝑥 + 2)𝑑𝑥 =

2

9∫

1

(𝑥 − 1)𝑑𝑥 +

1

3∫

1

(𝑥 − 1)2𝑑𝑥 −

2

9∫

1

(𝑥 + 2)𝑑𝑥

=2

9log|𝑥 − 1| +

1

3(

−1

𝑥 − 1) −

2

9log|𝑥 + 2| + C

=2

9log |

𝑥−1

𝑥+2| −

1

3(𝑥−1)+ C


(𝑥−1)2(𝑥+2)=

2

9log |

𝑥−1

𝑥+2| −

1

3(𝑥−1)+ C [3

Marks]

9. Integrate the rational function 3𝑥+5

𝑥3−𝑥2−𝑥+1 [6

Marks]

Solution:



We need to integrate 3𝑥+5

𝑥3−𝑥2−𝑥+1

3𝑥+5

(𝑥−1)2(𝑥+1)=

𝐴

(𝑥−1)+

𝐵

(𝑥−1)2 +𝐶

(𝑥+1) [

𝟏

𝟐

Mark]

⇒ 3𝑥 + 5 = 𝐴(𝑥 − 1)(𝑥 + 1) + 𝐵(𝑥 + 1) + 𝐶(𝑥 − 1)2

⇒ 3𝑥 + 5 = 𝐴(𝑥2 − 1) + 𝐵(𝑥 + 1) + 𝐶(𝑥2 + 1 − 2𝑥) … (1)


𝐴 + 𝐶 = 0

𝐵 − 2𝐶 = 3

– 𝐴 + 𝐵 + 𝐶 = 5

On solving, we get

𝐵 = 4

𝐴 = −1

2 and 𝐶 =

1

2 [2

Marks]

∴3𝑥+5

(𝑥−1)2(𝑥+1)=

−1

2(𝑥−1)+

4

(𝑥−1)2 +1

2(𝑥+1) [

𝟏

𝟐

Mark]

⇒ ∫3𝑥 + 5

(𝑥 − 1)2(𝑥 + 1)𝑑𝑥 = −

1

2∫

1

𝑥 − 1𝑑𝑥 + 4 ∫

1

(𝑥 − 1)2𝑑𝑥 +

1

2∫

1

(𝑥 + 1)𝑑𝑥

= −1

2log|𝑥 − 1| + 4 (

−1

𝑥 − 1) +

1

2log|𝑥 + 1| + C

=1

2log |

𝑥+1

𝑥−1| −

4

𝑥−1+ C

Hence, Integration of 3𝑥+5

𝑥3−𝑥2−𝑥+1=

1

2log |

𝑥+1

𝑥−1| −

4

𝑥−1+ C [3

Marks]


(𝑥2−1)(2𝑥+3) [6

Marks]

Solution:

Given: 2𝑥−3

(𝑥2−1)(2𝑥+3)=

2𝑥−3

(𝑥+1)(𝑥−1)(2𝑥+3)



Let 2𝑥−3

(𝑥+1)(𝑥−1)(2𝑥+3)=

𝐴

(𝑥+1)+

𝐵

(𝑥−1)+

𝐶

(2𝑥+3) [

𝟏

𝟐

Mark]

⇒ (2𝑥 − 3) = 𝐴(𝑥 − 1)(2𝑥 + 3) + 𝐵(𝑥 + 1)(2𝑥 + 3) + 𝐶(𝑥 + 1)(𝑥 − 1)

⇒ (2𝑥 − 3) = 𝐴(2𝑥2 + 𝑥 − 3) + 𝐵(2𝑥2 + 5𝑥 + 3) + 𝐶(𝑥2 − 1)

⇒ (2𝑥 − 3) = (2𝐴 + 2𝐵 + 𝐶)𝑥2 + (𝐴 + 5𝐵)𝑥 + (−3𝐴 + 3𝐵 − 𝐶)

Equating the coefficients of 𝑥2, 𝑥 and constant, we get

2𝐴 + 2𝐵 + 𝐶 = 0

𝐴 + 5𝐵 = 2

– 3𝐴 + 3𝐵– 𝐶 =– 3

On solving, we get

𝐵 = −1

10, 𝐴 =

5

2, and 𝐶 = −

24

5 [2

Marks]

∴2𝑥−3

(𝑥+1)(𝑥−1)(2𝑥+3)=

5

2(𝑥+1)−

1

10(𝑥−1)−

24

5(2𝑥+3) [

𝟏

𝟐

Mark]

⇒ ∫2𝑥 − 3

(𝑥2 − 1)(2𝑥 + 3)𝑑𝑥 =

5

2∫

1

(𝑥 + 1)𝑑𝑥 −

1

10∫

1

𝑥 − 1𝑑𝑥 −

24

5∫

1

(2𝑥 + 3)𝑑𝑥

=5

2log|𝑥 + 1| −

1

10log|𝑥 − 1| −

24

5 × 2log|2𝑥 + 3|

=5

2log|𝑥 + 1| −

1

10log|𝑥 − 1| −

12

5log|2𝑥 + 3| + C


(𝑥2−1)(2𝑥+3)=

5

2log|𝑥 + 1| −

1

10log|𝑥 − 1| −

12

5log|2𝑥 + 3| + C [3

Marks]


(𝑥+1)(𝑥2−4) [6

Marks]

Solution:


(𝑥+1)(𝑥2−4)=

5𝑥

(𝑥+1)(𝑥+2)(𝑥−2)



Let 5𝑥

(𝑥+1)(𝑥+2)(𝑥−2)=

𝐴

(𝑥+1)+

𝐵

(𝑥+2)+

𝐶

(𝑥−2) [

𝟏

𝟐

Mark]

⇒ 5𝑥 = 𝐴(𝑥 + 2)(𝑥 − 2) + 𝐵(𝑥 + 1)(𝑥 − 2) + 𝐶(𝑥 + 1)(𝑥 + 2) … (1)

Equating the coefficients of 𝑥2, 𝑥 and constant, we get

𝐴 + 𝐵 + 𝐶 = 0

−𝐵 + 3𝐶 = 5 and

−4𝐴 − 2𝐵 + 2𝐶 = 0

On solving, we get

𝐴 =5

3, 𝐵 = −

5

2, and 𝐶 =

5

6 [2

Marks]

∴5𝑥

(𝑥+1)(𝑥+2)(𝑥−2)=

5

3(𝑥+1)−

5

2(𝑥+2)+

5

6(𝑥−2) [

𝟏

𝟐

Mark]

⇒ ∫5𝑥

(𝑥 + 1)(𝑥2 − 4)𝑑𝑥

=5

3∫

1

(𝑥 + 1)𝑑𝑥 −

5

2∫

1

(𝑥 + 2)𝑑𝑥 +

5

6∫

1

(𝑥 − 2)𝑑𝑥

=5

3log|𝑥 + 1| −

5

2log|𝑥 + 2| +

5

6log|𝑥 − 2| + C

Hence, Integration of 5𝑥

(𝑥+1)(𝑥2−4)=

5

3log|𝑥 + 1| −

5

2log|𝑥 + 2| +

5

6log|𝑥 − 2| + C [3

Marks]

12. Integrate the rational function 𝑥3+𝑥+1

𝑥2−1 [4 Marks]

Solution:

We need to integrate 𝑥3+𝑥+1

𝑥2−1

It is seen that the given integrand is not a proper fraction.

Hence, on dividing (𝑥3 + 𝑥 + 1) by 𝑥2 − 1, we get

𝑥3 + 𝑥 + 1

𝑥2 − 1= 𝑥 +

2𝑥 + 1

𝑥2 − 1



Let 2𝑥+1

𝑥2−1=

𝐴

(𝑥+1)+

𝐵

(𝑥−1) [

𝟏

𝟐

Mark]

⇒ 2𝑥 + 1 = 𝐴(𝑥 − 1) + 𝐵(𝑥 + 1) … (1)

Equating the coefficients of 𝑥 and constant, we get

𝐴 + 𝐵 = 2

– 𝐴 + 𝐵 = 1

On solving, we get

𝐴 =1

2 and 𝐵 =

3

2 [1

Mark]

∴𝑥3+𝑥+1

𝑥2−1= 𝑥 +

1

2(𝑥+1)+

3

2(𝑥−1) [

𝟏

𝟐

Mark]

⇒ ∫𝑥3 + 𝑥 + 1

𝑥2 − 1𝑑𝑥 = ∫ 𝑥 𝑑𝑥 +

1

2∫

1

(𝑥 + 1)𝑑𝑥 +

3

2∫

1

(𝑥 − 1)𝑑𝑥

=𝑥2

2+

1

2log|𝑥 + 1| +

3

2log|𝑥 − 1| + C

Hence, Integration of 𝑥3+𝑥+1

𝑥2−1=

𝑥2

2+

1

2log|𝑥 + 1| +

3

2log|𝑥 − 1| + C [2

Marks]


(1−𝑥)(1+𝑥2) [6

Marks]

Solution:


(1−𝑥)(1+𝑥2)

Let 2

(1−𝑥)(1+𝑥2)=

𝐴

(1−𝑥)+

𝐵𝑥+𝐶

(1+𝑥2) [

𝟏

𝟐

Mark]

⇒ 2 = 𝐴(1 + 𝑥2) + (𝐵𝑥 + 𝐶)(1 − 𝑥)

⇒ 2 = 𝐴 + 𝐴𝑥2 + 𝐵𝑥 − 𝐵𝑥2 + 𝐶 − 𝐶𝑥

Equating the coefficient of 𝑥2, 𝑥, and constant term, we get

𝐴 − 𝐵 = 0



𝐵 − 𝐶 = 0

𝐴 + 𝐶 = 0


𝐴 = 1, 𝐵 = 1, and 𝐶 = 1 [2

Marks]

∴2

(1−𝑥)(1+𝑥2)=

1

1−𝑥+

𝑥+1

1+𝑥2 [𝟏

𝟐

Mark]

⇒ ∫2

(1 − 𝑥)(1 + 𝑥2)𝑑𝑥 = ∫

1

1 − 𝑥𝑑𝑥 + ∫

𝑥

1 + 𝑥2𝑑𝑥 + ∫

1

1 + 𝑥2𝑑𝑥

= − ∫1

𝑥 − 1𝑑𝑥 +

1

2∫

2𝑥

1 + 𝑥2𝑑𝑥 + ∫

1

1 + 𝑥2𝑑𝑥

= − log|𝑥 − 1| +1

2log|1 + 𝑥2| + tan−1𝑥 + 𝐶


(1−𝑥)(1+𝑥2)= − log|𝑥 − 1| +

1

2log|1 + 𝑥2| + tan−1𝑥 + 𝐶 [3

Marks]


(𝑥+2)2 [4 Marks]

Solution:

We need to integrate 3𝑥−1

(𝑥+2)2

Let 3𝑥−1

(𝑥+2)2 =𝐴

(𝑥+2)+

𝐵

(𝑥+2)2 [𝟏

𝟐

Mark]

⇒ 3𝑥 − 1 = 𝐴(𝑥 + 2) + 𝐵

Equating the coefficient of 𝑥 and constant term, we get

𝐴 = 3

2𝐴 + 𝐵 = −1 ⇒ 𝐵 = −7 [1

Mark]

∴3𝑥−1

(𝑥+2)2 =3

(𝑥+2)−

7

(𝑥+2)2 [𝟏

𝟐

Mark]



⇒ ∫3𝑥 − 1

(𝑥 + 2)2𝑑𝑥 = 3 ∫

1

(𝑥 + 2)𝑑𝑥 − 7 ∫

1

(𝑥 + 2)2𝑑𝑥

= 3 log|𝑥 + 2| − 7 (−1

(𝑥 + 2)) + C


(𝑥+2)2 = 3 log|𝑥 + 2| +7

(𝑥+2)+ C [2

Marks]


(𝑥4−1) [6 Marks]

Solution:


(𝑥4−1)

1

(𝑥4 − 1)=

1

(𝑥2 − 1)(𝑥2 + 1)=

1

(𝑥 + 1)(𝑥 − 1)(1 + 𝑥2)

Let 1

(𝑥+1)(𝑥−1)(1+𝑥2)=

𝐴

(𝑥+1)+

𝐵

(𝑥−1)+

𝐶𝑥+𝐷

(𝑥2+1) [

𝟏

𝟐

Mark]

⇒ 1 = 𝐴(𝑥 − 1)(𝑥2 + 1) + 𝐵(𝑥 + 1)(𝑥2 + 1) + (𝐶𝑥 + 𝐷)(𝑥2 − 1)

⇒ 1 = 𝐴(𝑥3 + 𝑥 − 𝑥2 − 1) + 𝐵(𝑥3 + 𝑥 + 𝑥2 + 1) + 𝐶𝑥3 + 𝐷𝑥2 − 𝐶𝑥 − 𝐷

⇒ 1 = (𝐴 + 𝐵 + 𝐶)𝑥3 + (−𝐴 + 𝐵 + 𝐷)𝑥2 + (𝐴 + 𝐵 − 𝐶)𝑥 + (−𝐴 + 𝐵 − 𝐷)

Equating the coefficient of 𝑥3, 𝑥2, 𝑥, and constant term, we get

𝐴 + 𝐵 + 𝐶 = 0

−𝐴 + 𝐵 + 𝐷 = 0

𝐴 + 𝐵 − 𝐶 = 0

−𝐴 + 𝐵 − 𝐷 = 0


𝐴 = −1

4, 𝐵 =

1

4, 𝐶 = 0, and 𝐷 = −

1

2 [2

Marks]

∴1

𝑥4−1=

−1

4(𝑥+1)+

1

4(𝑥−1)−

1

2(𝑥2+1) [

𝟏

𝟐

Mark]



⇒ ∫1

𝑥4 − 1𝑑𝑥 = −

1

4log|𝑥 − 1| +

1

4log|𝑥 − 1| −

1

2tan−1𝑥 + 𝐶

=1

4log |

𝑥 − 1

𝑥 + 1| −

1

2tan−1𝑥 + C


(𝑥4−1)=

1

4log |

𝑥−1

𝑥+1| −

1

2tan−1𝑥 + C [3

Marks]


𝑥(𝑥𝑛+1) [6 Marks]

[Hint: multiply numerator and denominator by 𝑥𝑛−1 and put 𝑥𝑛 = 𝑡]

Solution:


𝑥(𝑥𝑛+1)

Multiplying numerator and denominator by 𝑥𝑛−1, we get

1

𝑥(𝑥𝑛 + 1)=

𝑥𝑛−1

𝑥𝑛−1𝑥(𝑥𝑛 + 1)=

𝑥𝑛−1

𝑥𝑛(𝑥𝑛 + 1)

Let 𝑥𝑛 = 𝑡 ⇒ 𝑛𝑥𝑛−1𝑑𝑥 = 𝑑𝑡 [𝟏

Mark]

∴ ∫1

𝑥(𝑥𝑛+1)𝑑𝑥 = ∫

𝑥𝑛−1

𝑥𝑛(𝑥𝑛+1)𝑑𝑥 =

1

𝑛∫

1

𝑡(𝑡+1)𝑑𝑡 [

𝟏

𝟐

Mark]

Let 1

𝑡(𝑡+1)=

𝐴

𝑡+

𝐵

(𝑡+1)

⇒ 1 = 𝐴(1 + 𝑡) + 𝐵𝑡 … (1)

Equating the coefficients of 𝑡 and constant, we get

𝐴 = 1 and 𝐵 = −1 [1

Mark]

∴1

𝑡(𝑡+1)=

1

𝑡−

1

(1+𝑡) [

𝟏

𝟐

Mark]

⇒ ∫1

𝑥(𝑥𝑛 + 1)𝑑𝑥 =

1

𝑛∫ {

1

𝑡−

1

(𝑡 + 1)} 𝑑𝑡

=1

𝑛[log|𝑡| − log|𝑡 + 1| + 𝐶



=1

𝑛[log|𝑥𝑛| − log|𝑥𝑛 + 1|] + 𝐶

=1

𝑛log |

𝑥𝑛

𝑥𝑛 + 1| + 𝐶


𝑥(𝑥𝑛+1)=

1

𝑛log |

𝑥𝑛

𝑥𝑛+1| + 𝐶 [3

Marks]

17. Integrate the rational function cos 𝑥

(1−sin 𝑥)(2−sin 𝑥)) [Hint: Put sin 𝑥 = 𝑡] [6

Marks]

Solution:


(1−sin 𝑥(2−sin 𝑥))

Let sin 𝑥 = 𝑡 ⇒ cos 𝑥 𝑑𝑥 = 𝑑𝑡 [𝟏

Mark]

∴ ∫cos 𝑥

(1 − sin 𝑥)(2 − sin 𝑥)𝑑𝑥 = ∫

𝑑𝑡

(1 − 𝑡)(2 − 𝑡)

Let 1

(1−𝑡)(2−𝑡)=

𝐴

1−𝑡+

𝐵

(2−𝑡) [

𝟏

𝟐

Mark]

⇒ 1 = 𝐴(2 − 𝑡) + 𝐵(1 − 𝑡) … (1)

Equating the coefficient of 𝑡 and constant, we get

−2𝐴 − 𝐵 = 0 and

2𝐴 + 𝐵 = 1

On solving, we get

𝐴 = 1 and 𝐵 = −1 [2

Marks]

∴1

(1−𝑡)(2−𝑡)=

1

(1−𝑡)−

1

(2−𝑡) [

𝟏

𝟐

Mark]

⇒ ∫cos𝑥

(1 − sin𝑥)(2 − sin𝑥)𝑑𝑥 = ∫ {

1

1 − 𝑡−

1

(2 − 𝑡)} 𝑑𝑡

= − log|1 − 𝑡| + log|2 − 𝑡| + C



= log |2 − 𝑡

1 − 𝑡| + C

= log |2 − sin𝑥

1 − sin𝑥| + C

Hence, integration of cos 𝑥

(1−sin 𝑥)(2−sin 𝑥))= log |

2−sin𝑥

1−sin𝑥| + C [2

Marks]

18. Integrate the rational function (𝑥2+1)(𝑥2+2)

(𝑥2+3)(𝑥2+4) [6

Marks]

Solution:

We need to integrate (𝑥2+1)(𝑥2+2)

(𝑥2+3)(𝑥2+4)

(𝑥2 + 1)(𝑥2 + 2)

(𝑥2 + 3)(𝑥2 + 4)= 1 −

(4𝑥2 + 10)

(𝑥2 + 3)(𝑥2 + 4)

4𝑥2+10

(𝑥2+3)(𝑥2+4)=

𝐴𝑥+𝐵

(𝑥2+3)+

𝐶𝑥+𝐷

(𝑥2+4) [

𝟏

𝟐

Mark]

⇒ 4𝑥2 + 10 = (𝐴𝑥 + 𝐵)(𝑥2 + 4) + (𝐶𝑥 + 𝐷)(𝑥2 + 3)

⇒ 4𝑥2 + 10 = 𝐴𝑥3 + 4𝐴𝑥 + 𝐵𝑥2 + 4𝐵 + 𝐶𝑥3 + 3𝐶𝑥 + 𝐷𝑥2 + 3𝐷

⇒ 4𝑥2 + 10 = (𝐴 + 𝐶)𝑥3 + (𝐵 + 𝐷)𝑥2 + (4𝐴 + 3𝐶)𝑥 + (4𝐵 + 3𝐷) [1

Mark]

Equating the coefficients of 𝑥3, 𝑥2, 𝑥, and constant term, we get

𝐴 + 𝐶 = 0

𝐵 + 𝐷 = 4

4𝐴 + 3𝐶 = 0

4𝐵 + 3𝐷 = 10


𝐴 = 0, 𝐵 = −2, 𝐶 = 0, and 𝐷 = 6 [2

Marks]



∴4𝑥2+10

(𝑥2+3)(𝑥2+4)=

−2

(𝑥2+3)+

6

(𝑥2+4) [

𝟏

𝟐

Mark]

Hence, (𝑥2+1)(𝑥2+2)

(𝑥2+3)(𝑥2+4)= 1 − (

−2

(𝑥2+3)+

6

(𝑥2+4))

⇒ ∫(𝑥2 + 1)(𝑥2 + 2)

(𝑥2 + 3)(𝑥2 + 4)𝑑𝑥 = ∫ {1 +

2

(𝑥2 + 3)−

6

(𝑥2 + 4)} 𝑑𝑥

= ∫ {1 +2

𝑥2 + (√3)2 −

6

𝑥2 + 22} 𝑑𝑥

= 𝑥 + 2 (1

√3tan−1

𝑥

√3) − 6 (

1

2tan−1

𝑥

2) + C

= 𝑥 +2

√3tan−1

𝑥

√3− 3tan−1

𝑥

2+ C

Hence, integration of (𝑥2+1)(𝑥2+2)

(𝑥2+3)(𝑥2+4)= 𝑥 +

2

√3tan−1 𝑥

√3− 3tan−1 𝑥

2+ C [2

Marks]


(𝑥2+1)(𝑥2+3) [6

Marks]

Solution:


(𝑥2+1)(𝑥2+3)

Let 𝑥2 = 𝑡 ⇒ 2𝑥 𝑑𝑥 = 𝑑𝑡

∴ ∫2𝑥

(𝑥2+1)(𝑥2+3)𝑑𝑥 = ∫

𝑑𝑡

(𝑡+1)(𝑡+3) … (1) [1

Mark]

Let 1

(𝑡+1)(𝑡+3)=

𝐴

(𝑡+1)+

𝐵

(𝑡+3) [

𝟏

𝟐

Mark]

⇒ 1 = 𝐴(𝑡 + 3) + 𝐵(𝑡 + 1)


𝐴 + 𝐵 = 0 and 3𝐴 + 𝐵 = 1

On solving, we get



𝐴 =1

2 and 𝐵 = −

1

2 [2

Marks]

∴1

(𝑡+1)(𝑡+3)=

1

2(𝑡+1)−

1

2(𝑡+3) [

𝟏

𝟐

Mark]

⇒ ∫2𝑥

(𝑥2 + 1)(𝑥2 + 3)𝑑𝑥 = ∫ {

1

2(𝑡 + 1)−

1

2(𝑡 + 3)} 𝑑𝑡

=1

2log|𝑡 + 1| −

1

2log|𝑡 + 3| + 𝐶

=1

2log |

𝑡 + 1

𝑡 + 3| + 𝐶

=1

2log |

𝑥2 + 1

𝑥2 + 3| + 𝐶

Hence, integration of 2𝑥

(𝑥2+1)(𝑥2+3)=

1

2log |

𝑥2+1

𝑥2+3| + 𝐶 [2

Marks]


𝑥(𝑥4−1) [6 Marks]

Solution:


𝑥(𝑥4−1)

Multiplying numerator and denominator by 𝑥3, we get

1

𝑥(𝑥4 − 1)=

𝑥3

𝑥4(𝑥4 − 1)

∴ ∫1

𝑥(𝑥4 − 1)𝑑𝑥 = ∫

𝑥3

𝑥4(𝑥4 − 1)𝑑𝑥

Let 𝑥4 = 𝑡 ⇒ 4𝑥3𝑑𝑥 = 𝑑𝑡 [1

Mark]

∴ ∫1

𝑥(𝑥4 − 1)𝑑𝑥 =

1

4∫

𝑑𝑡

𝑡(𝑡 − 1)

Let 1

𝑡(𝑡−1)=

𝐴

𝑡+

𝐵

(𝑡−1)



⇒ 1 = 𝐴(𝑡 − 1) + 𝐵𝑡 … (1) [𝟏

𝟐

Mark]


𝐴 = −1 and 𝐵 = 1 [1

Mark]

⇒1

𝑡(𝑡+1)=

−1

𝑡+

1

𝑡−1 [

𝟏

𝟐

Mark]

⇒ ∫1

𝑥(𝑥4 − 1)𝑑𝑥 =

1

4∫ {

−1

𝑡+

1

𝑡 − 1} 𝑑𝑡

=1

4[− log|𝑡| + log|𝑡 − 1|] + 𝐶

=1

4log |

𝑡 − 1

𝑡| + C

=1

4log |

𝑥4 − 1

𝑥4| + C


𝑥(𝑥4−1)=

1

4log |

𝑥4−1

𝑥4 | + C [3

Marks]


(𝑒𝑥−1) [Hint: Put 𝑒𝑥 = 𝑡] [4 Marks]

Solution:


(𝑒𝑥−1)

Let 𝑒𝑥 = 𝑡 ⇒ 𝑒𝑥𝑑𝑥 = 𝑑𝑡 [𝟏

𝟐

Mark]

⇒ ∫1

𝑒𝑥 − 1𝑑𝑥 = ∫

1

𝑡 − 1×

𝑑𝑡

𝑡= ∫

1

𝑡(𝑡 − 1)𝑑𝑡

Let 1

𝑡(𝑡−1)=

𝐴

𝑡+

𝐵

𝑡−1

⇒ 1 = 𝐴(𝑡 − 1) + 𝐵𝑡 . . . (1)

Equating the coefficient of 𝑡 and constant, we get



𝐴 = −1 and 𝐵 = 1 [1

Mark]

∴1

𝑡(𝑡−1)=

−1

𝑡+

1

𝑡−1 [

𝟏

𝟐

Mark]

⇒ ∫1

𝑡(𝑡 − 1)𝑑𝑡 = ∫ {

−1

𝑡+

1

𝑡 − 1} 𝑑𝑡

= − log|𝑡| + log|𝑡 − 1| + 𝐶

= log |𝑡 − 1

𝑡| + 𝐶

= log |𝑒𝑥 − 1

𝑒𝑥 | + 𝐶


(𝑒𝑥−1)= log |

𝑒𝑥−1

𝑒𝑥 | + 𝐶 [2

Marks]

22. ∫𝑥 𝑑𝑥

(𝑥−1)(𝑥−2) equals [4

Marks]

(A) log |(𝑥−1)2

𝑥−2| + C

(B) log |(𝑥−2)2

𝑥−1| + C

(C) log |(𝑥−1

𝑥−2)

2| + C

(D) log|(𝑥 − 1)(𝑥 − 2)| + C

Solution:

We need to find ∫𝑥 𝑑𝑥

(𝑥−1)(𝑥−2)

Let 𝑥

(𝑥−1)(𝑥−2)=

𝐴

(𝑥−1)+

𝐵

(𝑥−2)

⇒ 𝑥 = 𝐴(𝑥 − 2) + 𝐵(𝑥 − 1) … (1) [𝟏

𝟐

Mark]

Equating the coefficient of 𝑥 and constant, we get



𝐴 = −1, and 𝐵 = 2 [1

Mark]

∴𝑥

(𝑥−1)(𝑥−2)= −

1

(𝑥−1)+

2

(𝑥−2) [

𝟏

𝟐

Mark]

⇒ ∫𝑥

(𝑥 − 1)(𝑥 − 2)𝑑𝑥 = ∫ {

−1

(𝑥 − 1)+

2

(𝑥 − 2)} 𝑑𝑥

= − log|𝑥 − 1| + 2 log|𝑥 − 2| + C

= log |(𝑥 − 2)2

𝑥 − 1| + C

Hence, the correct answer is B. [2

Marks]

23. ∫𝑑𝑥

𝑥(𝑥2+1) equals [4 Marks]

(A) log|𝑥| −1

2log(𝑥2 + 1) + C

(B) log|𝑥| +1

2log(𝑥2 + 1) + C

(C) − log|𝑥| +1

2log(𝑥2 + 1) + C

(D) 1

2log|𝑥| + log(𝑥2 + 1) + C

Solution:

We need to find ∫𝑑𝑥

𝑥(𝑥2+1)

Let 1

𝑥(𝑥2+1)=

𝐴

𝑥+

𝐵𝑥+𝐶

𝑥2+1

⇒ 1 = 𝐴(𝑥2 + 1) + (𝐵𝑥 + 𝐶)𝑥 [𝟏

𝟐

Mark]

Equating the coefficients of 𝑥2, 𝑥, and constant term, we get

𝐴 + 𝐵 = 0

𝐶 = 0

𝐴 = 1




𝐴 = 1, 𝐵 = −1, and 𝐶 = 0 [1

Mark]

∴1

𝑥(𝑥2+1)=

1

𝑥+

−𝑥

𝑥2+1 [

𝟏

𝟐

Mark]

⇒ ∫1

𝑥(𝑥2 + 1)𝑑𝑥 = ∫ {

1

𝑥−

𝑥

𝑥2 + 1} 𝑑𝑥

= log|𝑥| −1

2log|𝑥2 + 1| + C

Hence, the correct answer is A [2

Marks]

Exercise 7.6

1. Integrate the function 𝑥 sin 𝑥 [2

Marks]

Solution:

We need to integrate 𝑥 sin 𝑥

Let 𝐼 = ∫ 𝑥 sin 𝑥 𝑑𝑥

Taking 𝑥 as first function and sin 𝑥 as second function and integrating using by parts, we get

𝐼 = 𝑥 ∫ sin 𝑥 𝑑𝑥 − ∫ {(𝑑

𝑑𝑥𝑥) ∫ sin 𝑥 𝑑𝑥} 𝑑𝑥

[∵ ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 = 𝑓(𝑥)∫ 𝑔(𝑥)𝑑𝑥 − ∫[𝑓′(𝑥)∫ 𝑔(𝑥)𝑑𝑥]𝑑𝑥] [1

Mark]

= 𝑥(− cos 𝑥) − ∫ 1 · (− cos 𝑥)𝑑𝑥

= −𝑥 cos 𝑥 + sin 𝑥 + C

Hence, integration of 𝑥 sin 𝑥 = −𝑥 cos 𝑥 + sin 𝑥 + 𝐶 [1

Mark]

2. Integrate the function 𝑥 sin 3𝑥. [2 Marks]



Solution:

We need to integrate 𝑥 sin 3𝑥

Let 𝐼 = ∫ 𝑥 sin 3𝑥 𝑑𝑥

Taking 𝑥 as first function and sin 3𝑥 as second function and integrating using by parts, we get

𝐼 = 𝑥 ∫ sin 3𝑥 𝑑𝑥 − ∫ {(𝑑

𝑑𝑥𝑥) ∫ sin3𝑥𝑑𝑥} 𝑑𝑥


Mark]

= 𝑥 (−cos3𝑥

3) − ∫ 1 · (

−cos3𝑥

3) 𝑑𝑥

=−𝑥 cos 3𝑥

3+

1

3∫ cos 3𝑥 𝑑𝑥

=−𝑥 cos 3𝑥

3+

1

9sin 3𝑥 + C

Hence, Integration of 𝑥 sin 3𝑥 =−𝑥 cos 3𝑥

3+

1

9sin 3𝑥 + C [1

Mark]

3. Integrate the function 𝑥2𝑒𝑥 [4

Marks]

Solution:

We need to integrate 𝑥2𝑒𝑥

Let 𝐼 = ∫ 𝑥2𝑒𝑥 𝑑𝑥

Taking 𝑥2 as first function and 𝑒𝑥 as second function and integrating using by parts, we get

𝐼 = 𝑥2 ∫ 𝑒𝑥𝑑𝑥 − ∫ {(𝑑

𝑑𝑥𝑥2) ∫ 𝑒𝑥𝑑𝑥} 𝑑𝑥 [∵ ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 = 𝑓(𝑥)∫ 𝑔(𝑥)𝑑𝑥 −

∫[𝑓′(𝑥)∫ 𝑔(𝑥)𝑑𝑥]𝑑𝑥]

[1 Mark]

= 𝑥2𝑒𝑥 − ∫ 2𝑥 · 𝑒𝑥𝑑𝑥

= 𝑥2𝑒𝑥 − 2 ∫ 𝑥 · 𝑒𝑥 𝑑𝑥 [1

Mark]



Again, integrating using by parts, we get

= 𝑥2𝑒𝑥 − 2 [𝑥 · ∫ 𝑒𝑥𝑑𝑥 − ∫ {(𝑑

𝑑𝑥𝑥) · ∫ 𝑒𝑥𝑑𝑥} 𝑑𝑥] [1

Mark]

= 𝑥2𝑒𝑥 − 2[𝑥𝑒𝑥 − ∫ 𝑒𝑥 𝑑𝑥]

= 𝑥2𝑒𝑥 − 2[𝑥𝑒𝑥 − 𝑒𝑥] + 𝐶

= 𝑥2𝑒𝑥 − 2𝑥𝑒𝑥 + 2𝑒𝑥 + C

= 𝑒𝑥(𝑥2 − 2𝑥 + 2) + C

Hence, integration of 𝑥2𝑒𝑥 = 𝑒𝑥(𝑥2 − 2𝑥 + 2) + C [1

Mark]

4. Integrate the function 𝑥 log 𝑥 [2

Marks]

Solution:

We need to integrate 𝑥 log 𝑥

Let 𝐼 = ∫ 𝑥 log 𝑥 𝑑𝑥

Taking log 𝑥 as first function and 𝑥 as second function and integrating using by parts, we get,

𝐼 = log𝑥 ∫ 𝑥 𝑑𝑥 − ∫ {(𝑑

𝑑𝑥log𝑥) ∫ 𝑥 𝑑𝑥} 𝑑𝑥


Mark]

= log𝑥 ·𝑥2

2− ∫

1

𝑥·

𝑥2

2𝑑𝑥

=𝑥2 log 𝑥

2− ∫

𝑥

2𝑑𝑥

=𝑥2log𝑥

2−

𝑥2

4+ C

Hence, integration of 𝑥 log 𝑥 =𝑥2log𝑥

2−

𝑥2

4+ C [1

Mark]

5. Integrate the function 𝑥 log 2𝑥 [2 Marks]



Solution:

We need to integrate 𝑥 log 2𝑥

Let 𝐼 = ∫ 𝑥 log 2𝑥 𝑑𝑥

Taking log 2𝑥 as first function and 𝑥 as second function and integrating using by parts, we get,

𝐼 = log2𝑥 ∫ 𝑥 𝑑𝑥 − ∫ {(𝑑

𝑑𝑥2log𝑥) ∫ 𝑥 𝑑𝑥} 𝑑𝑥


Mark]

= log 2𝑥 ·𝑥2

2− ∫

2

2𝑥·

𝑥2

2𝑑𝑥

=𝑥2 log 2𝑥

2− ∫

𝑥

2𝑑𝑥

=𝑥2 log 2𝑥

2−

𝑥2

4+ C

Hence, integration of 𝑥 log 2𝑥 =𝑥2 log 2𝑥

2−

𝑥2

4+ C [1

Mark]

6. Integrate the function 𝑥2 log 𝑥 [2 Marks]

Solution:

We need to integrate 𝑥2 log 𝑥

Let 𝐼 = ∫ 𝑥2 log 𝑥 𝑑𝑥

Taking log 𝑥 as first function and 𝑥2 as second function and integrating using by parts, we get,

𝐼 = log𝑥 ∫ 𝑥2 𝑑𝑥 − ∫ {(𝑑

𝑑𝑥log 𝑥) ∫ 𝑥2 𝑑𝑥} 𝑑𝑥


Mark]

= log𝑥 (𝑥3

3) − ∫

1

𝑥·

𝑥3

3𝑑𝑥



=𝑥3 log 𝑥

3− ∫

𝑥2

3𝑑𝑥

=𝑥3 log 𝑥

3−

𝑥3

9+ C

Hence, integration of 𝑥2 log 𝑥 =𝑥3 log 𝑥

3−

𝑥3

9+ C [1

Mark]

7. Integrate the function 𝑥 sin−1 𝑥 [4

Marks]

Solution:

We need to integrate 𝑥 sin−1 𝑥

Let 𝐼 = ∫ 𝑥 sin−1 𝑥 𝑑𝑥

Taking sin−1 𝑥 as first function and 𝑥 as second function and integrating using by parts, we get,

𝐼 = sin−1𝑥 ∫ 𝑥 𝑑𝑥 − ∫ {(𝑑

𝑑𝑥sin−1𝑥) ∫ 𝑥 𝑑𝑥} 𝑑𝑥 [1

Mark]

[∵ ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 = 𝑓(𝑥)∫ 𝑔(𝑥)𝑑𝑥 − ∫[𝑓′(𝑥)∫ 𝑔(𝑥)𝑑𝑥]𝑑𝑥]

= sin−1𝑥 (𝑥2

2) − ∫

1

√1−𝑥2·

𝑥2

2𝑑𝑥

=𝑥2sin−1𝑥

2+

1

2∫

−𝑥2

√1−𝑥2𝑑𝑥

=𝑥2sin−1𝑥

2+

1

2∫ {

1−𝑥2

√1−𝑥2−

1

√1−𝑥2} 𝑑𝑥 [1

Mark]

=𝑥2sin−1𝑥

2+

1

2∫ {√1 − 𝑥2 −

1

√1−𝑥2} 𝑑𝑥

=𝑥2sin−1𝑥

2+

1

2{∫ √1 − 𝑥2 𝑑𝑥 − ∫

1

√1−𝑥2𝑑𝑥}

=𝑥2sin−1𝑥

2+

1

2{

𝑥

2√1 − 𝑥2 +

1

2sin−1𝑥 − sin−1𝑥} + 𝐶 [1

Mark]

=𝑥2sin−1𝑥

2+

𝑥

4√1 − 𝑥2 +

1

4sin−1𝑥 −

1

2sin−1𝑥 + 𝐶

=1

4(2𝑥2 − 1)sin−1𝑥 +

𝑥

4√1 − 𝑥2 + 𝐶



Hence, integration of 𝑥 sin−1 𝑥 =1

4(2𝑥2 − 1)sin−1𝑥 +

𝑥

4√1 − 𝑥2 + 𝐶 [1

Mark]

8. Integrate the function 𝑥 tan−1 𝑥 [4

Marks]

Solution:

We need to integrate 𝑥 tan−1 𝑥

Let 𝐼 = ∫ 𝑥 tan−1 𝑥 𝑑𝑥

Taking tan−1 𝑥 as first function and 𝑥 as second function and integrating using by parts, we get,

𝐼 = tan−1𝑥 ∫ 𝑥 𝑑𝑥 − ∫ {(𝑑

𝑑𝑥tan−1𝑥) ∫ 𝑥 𝑑𝑥} 𝑑𝑥 [1

Mark]


= tan−1𝑥 (𝑥2

2) − ∫

1

1 + 𝑥2·

𝑥2

2𝑑𝑥

=𝑥2tan−1𝑥

2−

1

2∫

𝑥2

1 + 𝑥2𝑑𝑥

=𝑥2tan−1𝑥

2−

1

2∫ (

𝑥2+1

1+𝑥2 −1

1+𝑥2) 𝑑𝑥 [1

Mark]

=𝑥2tan−1𝑥

2−

1

2∫ (1 −

1

1+𝑥2) 𝑑𝑥

=𝑥2tan−1𝑥

2−

1

2(𝑥 − tan−1𝑥) + C [1

Mark]

=𝑥2

2tan−1𝑥 −

𝑥

2+

1

2tan−1𝑥 + C

Hence, Integration of 𝑥 tan−1 𝑥 =𝑥2

2tan−1𝑥 −

𝑥

2+

1

2tan−1𝑥 + C [1

Mark]

9. Integrate the function 𝑥 cos−1 𝑥 [4

Marks]



Solution:

We need to integrate 𝑥 cos−1 𝑥

Let 𝐼 = ∫ 𝑥 cos−1 𝑥 𝑑𝑥

Taking cos−1 𝑥 as first function and 𝑥 as second function and integrating using by parts, we get,

𝐼 = cos−1𝑥 ∫ 𝑥 𝑑𝑥 − ∫ {(𝑑

𝑑𝑥cos−1𝑥) ∫ 𝑥 𝑑𝑥} 𝑑𝑥 [1

Mark]


= cos−1𝑥𝑥2

2− ∫

−1

√1 − 𝑥2·

𝑥2

2𝑑𝑥

=𝑥2cos−1𝑥

2−

1

2∫

1 − 𝑥2 − 1

√1 − 𝑥2𝑑𝑥

=𝑥2cos−1𝑥

2−

1

2∫ {√1 − 𝑥2 + (

−1

√1−𝑥2)} 𝑑𝑥 [1

Mark]

=𝑥2cos−1𝑥

2−

1

2∫ √1 − 𝑥2 𝑑𝑥 −

1

2∫ (

−1

√1 − 𝑥2) 𝑑𝑥

=𝑥2cos−1𝑥

2−

1

2[

𝑥

2√1 − 𝑥2 +

1

2sin−1𝑥] −

1

2cos−1𝑥 + 𝐶 [1

Mark]

=𝑥2cos−1𝑥

2−

1

2[𝑥

2√1 − 𝑥2 −

1

2cos−1𝑥] −

1

2cos−1𝑥 + 𝐶1

=(2𝑥2 − 1)

4cos−1𝑥 −

𝑥

4√1 − 𝑥2 + 𝐶1

Hence, integration of 𝑥 cos−1 𝑥 =(2𝑥2−1)

4cos−1𝑥 −

𝑥

4√1 − 𝑥2 + 𝐶1 [1

Mark]

10. Integrate the function (sin−1 𝑥)2 [4

Marks]

Solution:

We need to integrate (sin−1 𝑥)2

Let 𝐼 = ∫(sin−1 𝑥)2 ∙ 1 𝑑𝑥



Taking (sin−1 𝑥)2 as first function and 1 as second function and integrating using by parts, we

get,

𝐼 = (sin−1𝑥) ∫ 1 𝑑𝑥 − ∫ {𝑑

𝑑𝑥(sin−1𝑥)2 · ∫ 1 · 𝑑𝑥} 𝑑𝑥 [1

Mark]


= (sin−1𝑥)2 · 𝑥 − ∫2sin−1𝑥

√1 − 𝑥2· 𝑥𝑑𝑥

= 𝑥(sin−1𝑥)2 + ∫ sin−1 𝑥 · (−2𝑥

√1 − 𝑥2) 𝑑𝑥

= 𝑥(sin−1𝑥)2 + [sin−1𝑥 ∫−2𝑥

√1−𝑥2𝑑𝑥 − ∫ {(

𝑑

𝑑𝑥sin−1𝑥) ∫

−2𝑥

√1−𝑥2𝑑𝑥} 𝑑𝑥]

= 𝑥(sin−1𝑥)2 + [sin−1𝑥 · 2√1 − 𝑥2 − ∫1

√1−𝑥2· 2√1 − 𝑥2𝑑𝑥] [2

Marks]

= 𝑥(sin−1𝑥)2 + 2√1 − 𝑥2sin−1𝑥 − ∫ 2 𝑑𝑥

= 𝑥(sin−1𝑥)2 + 2√1 − 𝑥2sin−1𝑥 − 2𝑥 + 𝐶

Hence, integration of (sin−1 𝑥)2 = 𝑥(sin−1𝑥)2 + 2√1 − 𝑥2sin−1𝑥 − 2𝑥 + 𝐶 [1

Mark]

11. Integrate the function 𝑥 cos−1 𝑥

√1−𝑥2 [4 Marks]

Solution:

We need to integrate 𝑥 cos−1 𝑥

√1−𝑥2

Let 𝐼 = ∫𝑥 cos−1 𝑥


𝐼 =−1

2∫

−2𝑥

√1 − 𝑥2· cos−1𝑥𝑑𝑥

Taking cos−1 𝑥 as first function and (−2𝑥

√1−𝑥2) as second function and integrating using by parts,

we get,

𝐼 =−1

2[cos−1𝑥 ∫

−2𝑥

√1−𝑥2𝑑𝑥 − ∫ {(

𝑑

𝑑𝑥cos−1𝑥) ∫

−2𝑥

√1−𝑥2𝑑𝑥} 𝑑𝑥] [1

Mark]




=−1

2[cos−1𝑥 · 2√1 − 𝑥2 − ∫

−1

√1−𝑥2· 2√1 − 𝑥2𝑑𝑥] [1

Mark]

=−1

2[2√1 − 𝑥2cos−1𝑥 + ∫ 2 𝑑𝑥]

=−1

2[2√1 − 𝑥2cos−1𝑥 + 2𝑥] + C

= − [√1 − 𝑥2cos−1𝑥 + 𝑥] + C

Hence, integration of 𝑥 cos−1 𝑥

√1−𝑥2= −[√1 − 𝑥2cos−1𝑥 + 𝑥] + C [2

Marks]

12. Integrate the function 𝑥 sec2 𝑥 [2 Marks]

Solution:

Let 𝐼 = ∫ 𝑥 sec2 𝑥𝑑𝑥

Taking 𝑥 as first function and sec2 𝑥 as second function and integrating using by parts, we get,

𝐼 = 𝑥 ∫ sec2 𝑥 𝑑𝑥 − ∫ {{𝑑

𝑑𝑥𝑥} ∫ sec2𝑥 𝑑𝑥} 𝑑𝑥 [1

Mark]


= 𝑥 tan 𝑥 − ∫ 1 · tan 𝑥 𝑑𝑥

= 𝑥 tan 𝑥 + log|cos𝑥| + C

Hence, integration of 𝑥 sec2 𝑥 = 𝑥 tan 𝑥 + log|cos𝑥| + 𝐶 [1

Mark]

13. Integrate the function tan−1 𝑥 [2 Marks]

Solution:



We need to integrate tan−1 𝑥

Let 𝐼 = ∫ 1 ∙ tan−1 𝑥𝑑𝑥

Taking tan−1 𝑥 as first function and 1 as second function and integrating using by parts, we get,

𝐼 = tan−1𝑥 ∫ 1 𝑑𝑥 − ∫ {(𝑑

𝑑𝑥tan−1𝑥) ∫ 1 · 𝑑𝑥} 𝑑𝑥 [1

Mark]


= tan−1𝑥 · 𝑥 − ∫1

1 + 𝑥2· 𝑥𝑑𝑥

= 𝑥tan−1𝑥 −1

2∫

2𝑥

1 + 𝑥2𝑑𝑥


2log|1 + 𝑥2| + C


2log(1 + 𝑥2) + C

Hence, integration of tan−1 𝑥 = 𝑥tan−1𝑥 −1

2log(1 + 𝑥2) + 𝐶 [1

Mark]

14. Integrate the function 𝑥 (log 𝑥)2 [4

Marks]

Solution:

We need to integrate 𝑥 (log 𝑥)2

𝐼 = ∫ 𝑥 (log 𝑥)2 𝑑𝑥

Taking (log 𝑥)2 as first function and 𝑥 as second function and integrating using by parts, we get

𝐼 = (log𝑥)2∫ 𝑥𝑑𝑥 − ∫ [{(𝑑

𝑑𝑥log𝑥)

2} ∫ 𝑥𝑑𝑥] 𝑑𝑥 [1

Mark]


=𝑥2

2(log 𝑥)2 − [∫ 2 log 𝑥 ·

1

𝑥·

𝑥2

2𝑑𝑥]



=𝑥2

2(log 𝑥)2 − ∫ 𝑥 log 𝑥 𝑑𝑥 [1

Mark]


𝐼 =𝑥2

2(log 𝑥)2 − [log 𝑥 ∫ 𝑥 𝑑𝑥 − ∫ {(

𝑑

𝑑𝑥log𝑥) ∫ 𝑥 𝑑𝑥} 𝑑𝑥] [1

Mark]


=𝑥2

2(log 𝑥)2 − [

𝑥2

2− log 𝑥 − ∫

1

𝑥·

𝑥2

2𝑑𝑥]

=𝑥2

2(log 𝑥)2 −

𝑥2

2log 𝑥 +

1

2∫ 𝑥 𝑑𝑥

=𝑥2

2(log 𝑥)2 −

𝑥2

2log 𝑥 +

𝑥2

4+ C

Hence, integration of 𝑥 (log 𝑥)2 =𝑥2

2(log 𝑥)2 −

𝑥2

2log 𝑥 +

𝑥2

4+ 𝐶 [1

Mark]

15. Integrate the function (𝑥2 + 1) log 𝑥 [6 Marks]

Solution:

We need to integrate (𝑥2 + 1) log 𝑥

Let 𝐼 = ∫(𝑥2 + 1) log 𝑥 𝑑𝑥 = ∫ 𝑥2 log 𝑥 𝑑𝑥 + ∫ log 𝑥 𝑑𝑥

𝐼 = 𝐼1 + 𝐼2 … (1) [1

Mark]

𝐼1 = ∫ 𝑥2 log 𝑥 𝑑𝑥 and 𝐼2 = ∫ log 𝑥 𝑑𝑥

𝐼1 = ∫ 𝑥2 log 𝑥 𝑑𝑥

Taking log 𝑥 as fist function and 𝑥2 as second function and integrating using by parts, we get

𝐼1 = log𝑥 − ∫ 𝑥2 𝑑𝑥 − ∫ {(𝑑

𝑑𝑥log 𝑥) ∫ 𝑥2 𝑑𝑥} 𝑑𝑥 [1

Mark]


= log𝑥 ·𝑥3

3− ∫

1

𝑥·

𝑥3

3𝑑𝑥



=𝑥3

3log𝑥 −

1

3(∫ 𝑥2 𝑑𝑥)

=𝑥3

3log𝑥 −

𝑥3

9+ 𝐶1 … (2) [1

Mark]

𝐼2 = ∫ log 𝑥 𝑑𝑥

Taking log 𝑥 as first function and 1 as second function and integrating using by parts, we get

𝐼2 = log 𝑥 ∫ 1 · 𝑑𝑥 − ∫ {(𝑑

𝑑𝑥log𝑥) ∫ 1 · 𝑑𝑥} [1

Mark]


= log𝑥 · 𝑥 − ∫1

𝑥· 𝑥𝑑𝑥

= 𝑥 log 𝑥 − ∫ 1 𝑑𝑥

= 𝑥 log 𝑥 − 𝑥 + 𝐶2 … (3) [1

Mark]

Using equations (2) and (3) in (1), we get

𝐼 =𝑥3

3log 𝑥 −

𝑥3

9+ C1 + 𝑥 log 𝑥 − 𝑥 + C2

=𝑥3

3log 𝑥 −

𝑥3

9+ 𝑥 log 𝑥 − 𝑥 + (C1 + C2)

= (𝑥3

3+ 𝑥) log 𝑥 −

𝑥3

9− 𝑥 + C

Hence, integration of (𝑥2 + 1) log 𝑥 = (𝑥3

3+ 𝑥) log 𝑥 −

𝑥3

9− 𝑥 + 𝐶 [1

Mark]

16. Integrate the function 𝑒𝑥(sin 𝑥 + cos 𝑥) [2

Marks]

Solution:

We need to integrate 𝑒𝑥(sin 𝑥 + cos 𝑥)



Let 𝐼 = ∫ 𝑒𝑥(sin 𝑥 + cos𝑥) 𝑑𝑥

Again, let 𝑓(𝑥) = sin 𝑥

Hence, 𝑓′(𝑥) = cos 𝑥

𝐼 = ∫ 𝑒𝑥{𝑓(𝑥) + 𝑓′(𝑥)} 𝑑𝑥 [1

Mark]

As we known that, ∫ 𝑒𝑥{𝑓(𝑥) + 𝑓′(𝑥)}𝑑𝑥 = 𝑒𝑥𝑓(𝑥) + C

∴ 𝐼 = 𝑒𝑥 sin 𝑥 + 𝐶

Hence, integration of 𝑒𝑥(sin 𝑥 + cos 𝑥) = 𝑒𝑥 sin 𝑥 + 𝐶 [1

Mark]

17. Integrate the function 𝑥𝑒𝑥

(1+𝑥)2 [2 Marks]

Solution:

We need to integrate 𝑥𝑒𝑥

(1+𝑥)2

Let 𝐼 = ∫𝑥𝑒𝑥

(1+𝑥)2 𝑑𝑥 = ∫ 𝑒𝑥 {𝑥

(1+𝑥)2} 𝑑𝑥

= ∫ 𝑒𝑥 {1 + 𝑥 − 1

(1 + 𝑥)2 } 𝑑𝑥

= ∫ 𝑒𝑥 {1

1 + 𝑥−

1

(1 + 𝑥)2} 𝑑𝑥

Let 𝑓(𝑥) =1

1+𝑥

Hence, 𝑓′(𝑥) =−1

(1+𝑥)2

⇒ ∫𝑥𝑒𝑥

(1+𝑥)2 𝑑𝑥 = ∫ 𝑒𝑥{𝑓(𝑥) + 𝑓′(𝑥)} 𝑑𝑥 [1

Mark]

As we known that, ∫ 𝑒𝑥{𝑓(𝑥) + 𝑓′(𝑥)} 𝑑𝑥 = 𝑒𝑥𝑓(𝑥) + C

∴ ∫𝑥𝑒𝑥

(1 + 𝑥)2𝑑𝑥 =

𝑒𝑥

1 + 𝑥+ C

Hence, integration of 𝑥𝑒𝑥

(1+𝑥)2 = ∫𝑥𝑒𝑥

(1+𝑥)2 𝑑𝑥 =𝑒𝑥

1+𝑥+ 𝐶 [1

Mark]



18. Integrate the function 𝑒𝑥 (1+sin𝑥

1+cos𝑥) [4

Marks]

Solution:

We need to integrate 𝑒𝑥 (1+sin 𝑥

1+cos 𝑥)

= 𝑒𝑥 (sin2𝑥

2+cos2𝑥

2+2sin

𝑥

2cos

𝑥

2

2 cos2𝑥

2

)

=𝑒𝑥(sin

𝑥

2+cos

𝑥

2)

2

2cos2𝑥

2

=1

2𝑒𝑥 · (

sin𝑥

2+cos

𝑥

2

cos𝑥

2

)

2

=1

2𝑒𝑥 [tan

𝑥

2+ 1]

2

=1

2𝑒𝑥 (1 + tan

𝑥

2)

2

=1

2𝑒𝑥 [1 + tan2 𝑥

2+ 2tan

𝑥

2]

=1

2𝑒𝑥 [sec2 𝑥

2+ 2tan

𝑥

2]

Hence, ∫𝑒𝑥(1+sin𝑥)

(1+cos𝑥)𝑑𝑥 = ∫ 𝑒𝑥 [

1

2sec2 𝑥

2+ tan

𝑥

2] 𝑑𝑥 ... (1) [2

Marks]

Let tan𝑥

2= 𝑓(𝑥)

Hence, 𝑓′(𝑥) =1

2sec2 𝑥

2

As we known that, ∫ 𝑒𝑥{𝑓(𝑥) + 𝑓′(𝑥)} 𝑑𝑥 = 𝑒𝑥𝑓(𝑥) + 𝐶 [1

Mark]

From equation (1) we get

∫𝑒𝑥(1+sin𝑥)

(1+cos𝑥)𝑑𝑥 = 𝑒𝑥tan

𝑥

2+ C

Hence, integration of 𝑒𝑥 (1+sin𝑥

1+cos𝑥) = 𝑒𝑥tan

𝑥

2+ C [1

Mark]



19. Integrate the function 𝑒𝑥 (1

𝑥−

1

𝑥2) [2

Marks]

Solution:

We need to integrate 𝑒𝑥 (1

𝑥−

1

𝑥2)

Let 𝐼 = ∫ 𝑒𝑥 [1

𝑥−

1

𝑥2] 𝑑𝑥

Also, let 1

𝑥= 𝑓(𝑥)


𝑥2

As we known that, ∫ 𝑒𝑥{𝑓(𝑥) + 𝑓′(𝑥)} 𝑑𝑥 = 𝑒𝑥𝑓(𝑥) + C [1

Mark]

∴ 𝐼 =𝑒𝑥

𝑥+ C

Hence, integration of 𝑒𝑥 (1

𝑥−

1

𝑥2) =𝑒𝑥

𝑥+ C [1

Mark]

20. Integrate the function (𝑥−3)𝑒𝑥

(𝑥−1)3 [2 Marks]

Solution:

We need to integrate (𝑥−3)𝑒𝑥

(𝑥−1)3

∫ 𝑒𝑥 {𝑥−3

(𝑥−1)3} 𝑑𝑥 = ∫ 𝑒𝑥 {𝑥−1−2

(𝑥−1)3} 𝑑𝑥

= ∫ 𝑒𝑥 {1

(𝑥−1)2 −2

(𝑥−1)3} 𝑑𝑥

𝑓(𝑥) =1

(𝑥−1)2


(𝑥−1)3 [1

Mark]

As we known that, ∫ 𝑒𝑥{𝑓(𝑥) + 𝑓′(𝑥)} 𝑑𝑥 = 𝑒𝑥𝑓(𝑥) + C

∴ ∫ 𝑒𝑥 {(𝑥−3)

(𝑥−1)2} 𝑑𝑥 =𝑒𝑥

(𝑥−1)2 + 𝐶



Hence, integration of (𝑥−3)𝑒𝑥

(𝑥−1)3 =𝑒𝑥

(𝑥−1)2 + 𝐶 [1

Mark]

21. Integrate the function 𝑒2𝑥sin 𝑥 [4 Marks]

Solution:

We need to integrate 𝑒2𝑥sin 𝑥

Let 𝐼 = ∫ 𝑒2𝑥 sin 𝑥 𝑑𝑥 … (1)

Integrating using by parts, we get

𝐼 = sin 𝑥 ∫ 𝑒2𝑥 𝑑𝑥 − ∫ {(𝑑

𝑑𝑥sin 𝑥) ∫ 𝑒2𝑥 𝑑𝑥} 𝑑𝑥 [1

Mark]

⇒ 𝐼 = sin 𝑥 ·𝑒2𝑥

2− ∫ cos 𝑥 ·

𝑒2𝑥

2𝑑𝑥

⇒ 𝐼 =𝑒2𝑥sin𝑥

2−

1

2∫ 𝑒2𝑥 cos 𝑥 𝑑𝑥


𝐼 =𝑒2𝑥·sin𝑥

2−

1

2[cos 𝑥 ∫ 𝑒2𝑥 𝑑𝑥 − ∫ {(

𝑑

𝑑𝑥cos 𝑥) ∫ 𝑒2𝑥 𝑑𝑥} 𝑑𝑥] [1

Mark]


⇒ 𝐼 =𝑒2𝑥 sin 𝑥

2−

1

2[cos 𝑥 ·

𝑒2𝑥

2− ∫(−sin𝑥)

𝑒2𝑥

2𝑑𝑥]

⇒ 𝐼 =𝑒2𝑥·sin 𝑥

2−

1

2[

𝑒2𝑥 cos 𝑥

2+

1

2∫ 𝑒2𝑥 sin 𝑥 𝑑𝑥]

⇒ 𝐼 =𝑒2𝑥sin𝑥

2−

𝑒2𝑥cos𝑥

4−

1

4𝐼 [From (1)]

⇒ 𝐼 +1

4𝐼 =

𝑒2𝑥 sin 𝑥

2−

𝑒2𝑥 cos 𝑥

4

⇒5

4𝐼 =

𝑒2𝑥sin𝑥

2−

𝑒2𝑥cos𝑥

4

⇒ 𝐼 =4

5[

𝑒2𝑥sin𝑥

2−

𝑒2𝑥cos𝑥

4] + C

⇒ 𝐼 =𝑒2𝑥

5[2sin𝑥 − cos𝑥] + C



Hence, integration of 𝑒2𝑥sin 𝑥 =𝑒2𝑥

5[2sin𝑥 − cos𝑥] + C [2

Marks]

22. Integrate the function sin−1 (2𝑥

1+𝑥2) [4

Marks]

Solution:

We need to integrate sin−1 (2𝑥

1+𝑥2)

Let 𝑥 = tan𝜃

Hence, 𝑑𝑥 = sec2𝜃 𝑑𝜃 [1

Mark]

∴ sin−1 (2𝑥

1 + 𝑥2) = sin−1 (

2tan𝜃

1 + tan2𝜃) = sin−1(sin2𝜃) = 2𝜃

∫ sin−1 (2𝑥

1+𝑥2) 𝑑𝑥 = ∫ 2𝜃 · sec2𝜃𝑑𝜃 = 2 ∫ 𝜃 · sec2𝜃𝑑𝜃 [1

Mark]

Integrating using by parts, we get

∫ sin−1 (2𝑥

1+𝑥2) 𝑑𝑥 = 2 [𝜃 · ∫ sec2 𝜃𝑑𝜃 − ∫ {(𝑑

𝑑𝜃𝜃) ∫ sec2 𝜃𝑑𝜃} 𝑑𝜃] [1

Mark]


= 2 [𝜃 · tan𝜃 − ∫ tan𝜃 𝑑𝜃]

= 2[𝜃tan𝜃 + log|𝑐𝑜𝑠𝜃|] + C

= 2 [𝑥tan𝜃 + log |1

√1 + 𝑥2|] + C

= 2𝑥tan−1𝑥 + log(1 + 𝑥2)12 + C

= 2𝑥tan−1𝑥 − log(1 + 𝑥2) + C

Hence, integration of sin−1 (2𝑥

1+𝑥2) = 2𝑥tan−1𝑥 − log(1 + 𝑥2) + C [1

Mark]



23. ∫ 𝑥2 𝑒𝑥3𝑑𝑥 equals [2

Marks]

(A) 1

3𝑒𝑥3

+ 𝐶

(B) 1

3𝑒𝑥2

+ 𝐶

(C) 1

2𝑒𝑥3

+ 𝐶

(D) 1

2𝑒𝑥2

+ 𝐶

Solution:

We need to integrate ∫ 𝑥2 𝑒𝑥3𝑑𝑥

Let 𝐼 = ∫ 𝑥2 𝑒𝑥3𝑑𝑥

Also, let 𝑥3 = 𝑡

Hence, 3𝑥2𝑑𝑥 = 𝑑𝑡

⇒ 𝐼 =1

3∫ 𝑒𝑡 𝑑𝑡 [1

Mark]

=1

3(𝑒𝑡) + C

=1

3𝑒𝑥3

+ C

Hence, the correct Answer is A [1

Mark]

24. ∫ 𝑒𝑥 sec𝑥(1 + tan𝑥)𝑑𝑥 equals [2 Marks]

(A) 𝑒𝑥cos𝑥 + 𝐶

(B) 𝑒𝑥sec𝑥 + 𝐶

(C) 𝑒𝑥sin𝑥 + 𝐶

(D) 𝑒𝑥tan𝑥 + 𝐶

Solution:

We need to integrate ∫ 𝑒𝑥 sec𝑥(1 + tan𝑥)𝑑𝑥



Let 𝐼 = ∫ 𝑒𝑥 sec𝑥(1 + tan𝑥)𝑑𝑥 = ∫ 𝑒𝑥 (sec𝑥 + sec𝑥 tan𝑥)𝑑𝑥

Also, let sec 𝑥 = 𝑓(𝑥)

Hence, 𝑓′(𝑥) = sec 𝑥 tan 𝑥 [1

Mark]

As we known that, ∫ 𝑒𝑥 {𝑓(𝑥) + 𝑓′(𝑥)}𝑑𝑥 = 𝑒𝑥𝑓(𝑥) + C

∴ 𝐼 = 𝑒𝑥sec𝑥 + C

Hence, the correct Answer is B [1

Mark]

Exercise 7.7

1. Integrate the function √4 − 𝑥2 [2 Marks]

Solution:

We need to integrate √4 − 𝑥2

Let 𝐼 = ∫ √4 − 𝑥2 𝑑𝑥 = ∫ √(2)2 − (𝑥)2𝑑𝑥

As we know that, ∫ √𝑎2 − 𝑥2𝑑𝑥 =𝑥

2√𝑎2 − 𝑥2 +

𝑎2

2sin−1 𝑥

𝑎+ 𝐶 [1

Mark]

Here, 𝑎 = 2, 𝑥 = 𝑥

∴ 𝐼 =𝑥

2√4 − 𝑥2 +

4

2sin−1

𝑥

2+ 𝐶

=𝑥

2√4 − 𝑥2 + 2 sin−1

𝑥

2+ 𝐶

Hence, integration of √4 − 𝑥2 =𝑥

2√4 − 𝑥2 + 2 sin−1 𝑥

2+ 𝐶 [1

Mark]

2. Integrate the function √1 − 4𝑥2 [2

Marks]



Solution:

We need to integrate √1 − 4𝑥2

Let 𝐼 = ∫ √1 − 4𝑥2 𝑑𝑥 = ∫ √(1)2 − (2𝑥)2𝑑𝑥

Let 2𝑥 = 𝑡 ⇒ 2𝑑𝑥 = 𝑑𝑡

∴ 𝐼 =1

2∫ √(1)2 − (𝑡)2𝑑𝑡 [1

Mark]

As we know that, ∫ √𝑎2 − 𝑥2 𝑑𝑥 =𝑥

2√𝑎2 − 𝑥2 +

𝑎2

2sin−1 𝑥

𝑎+ 𝐶

Here, 𝑎 = 1, 𝑥 = 𝑡

⇒ 𝐼 =1

2[

𝑡

2√1 − 𝑡2 +

1

2sin−1 𝑡] + 𝐶

=𝑡

4√1 − 𝑡2 +

1

4sin−1 𝑡 + 𝐶

=2𝑥

4√1 − 4𝑥2 +

1

4sin−1 2𝑥 + 𝐶 [substituting 𝑡]

=𝑥

2√1 − 4𝑥2 +

1

4sin−1 2𝑥 + 𝐶

Hence, integration of √1 − 4𝑥2 =𝑥

2√1 − 4𝑥2 +

1

4sin−1 2𝑥 + 𝐶 [1

Mark]

3. Integrate the function √𝑥2 + 4𝑥 + 6 [2

Marks]

Solution:

We need to integrate √𝑥2 + 4𝑥 + 6

Let 𝐼 = ∫ √𝑥2 + 4𝑥 + 6 𝑑𝑥

= ∫ √𝑥2 + 4𝑥 + 4 + 2 𝑑𝑥

= ∫ √(𝑥2 + 4𝑥 + 4) + 2 𝑑𝑥

= ∫ √(𝑥 + 2)2 + (√2)2

𝑑𝑥 [1

Mark]



As we know that, ∫ √𝑥2 + 𝑎2 𝑑𝑥 =𝑥

2√𝑥2 + 𝑎2 +

𝑎2

2log|𝑥 + √𝑥2 + 𝑎2| + 𝐶

Here, 𝑎 = √2, 𝑥 = 𝑥 + 2

∴ 𝐼 =(𝑥 + 2)

2√𝑥2 + 4𝑥 + 6 +

2

2log |(𝑥 + 2) + √𝑥2 + 4𝑥 + 6| + 𝐶

=(𝑥+2)

2√𝑥2 + 4𝑥 + 6 + log|(𝑥 + 2) + √𝑥2 + 4𝑥 + 6| + 𝐶 [1

Mark]

Hence, integration of √𝑥2 + 4𝑥 + 6 = (𝑥+2)

2√𝑥2 + 4𝑥 + 6 + log|(𝑥 + 2) + √𝑥2 + 4𝑥 + 6| + 𝐶

4. Integrate the function √𝑥2 + 4𝑥 + 1 [2

Marks]

Solution:

We need to integrate √𝑥2 + 4𝑥 + 1

Let 𝐼 = ∫ √𝑥2 + 4𝑥 + 1 𝑑𝑥

= ∫ √(𝑥2 + 4𝑥 + 4) − 3 𝑑𝑥

= ∫ √(𝑥 + 2)2 − (√3)2

𝑑𝑥 [1

Mark]

As we know that, ∫ √𝑥2 − 𝑎2𝑑𝑥 =𝑥

2√𝑥2 − 𝑎2 −

𝑎2

2log|𝑥 + √𝑥2 − 𝑎2| + 𝐶

Here, 𝑎 = √3, 𝑥 = 𝑥 + 2

∴ 𝐼 =(𝑥+2)

2√𝑥2 + 4𝑥 + 1 −

3

2log|(𝑥 + 2) + √𝑥2 + 4𝑥 + 1| + 𝐶 [1

Mark]

Hence, integration of √𝑥2 + 4𝑥 + 1 =(𝑥+2)

2√𝑥2 + 4𝑥 + 1 −

3

2log|(𝑥 + 2) + √𝑥2 + 4𝑥 + 1| +

𝐶

5. Integrate the function √1 − 4𝑥 − 𝑥2 [2

Marks]



Solution:

We need to integrate √1 − 4𝑥 − 𝑥2

Let 𝐼 = ∫ √1 − 4𝑥 − 𝑥2𝑑𝑥

= ∫ √1 − (𝑥2 + 4𝑥 + 4 − 4) 𝑑𝑥

= ∫ √1 + 4 − (𝑥 + 2)2 𝑑𝑥

= ∫ √(√5)2

− (𝑥 + 2)2 𝑑𝑥 [1

Mark]


2√𝑎2 − 𝑥2 +

𝑎2

2sin−1 𝑥

𝑎+ 𝐶

Here, 𝑎 = √5, 𝑥 = 𝑥 + 2

∴ 𝐼 =(𝑥 + 2)

2√1 − 4𝑥 − 𝑥2 +

5

2sin−1 (

𝑥 + 2

√5) + 𝐶

Hence, integration of √1 − 4𝑥 − 𝑥2 =(𝑥+2)

2√1 − 4𝑥 − 𝑥2 +

5

2sin−1 (

𝑥+2

√5) + 𝐶 [1

Mark]

6. Integrate the function √𝑥2 + 4𝑥 − 5 [2

Marks]

Solution:

We need to integrate √𝑥2 + 4𝑥 − 5

Let 𝐼 = ∫ √𝑥2 + 4𝑥 − 5 𝑑𝑥

= ∫ √(𝑥2 + 4𝑥 + 4) − 9 𝑑𝑥

= ∫ √(𝑥 + 2)2 − (3)2 𝑑𝑥 [1

Mark]

As we know that, ∫ √𝑥2 − 𝑎2𝑑𝑥 =𝑥

2√𝑥2 − 𝑎2 −

𝑎2

2log|𝑥 + √𝑥2 − 𝑎2| + 𝐶

Here, 𝑎 = 3, 𝑥 = 𝑥 + 2



∴ 𝐼 =(𝑥+2)

2√𝑥2 + 4𝑥 − 5 −

9

2log|(𝑥 + 2) + √𝑥2 + 4𝑥 − 5| + 𝐶 [1

Mark]

Hence, integration of √𝑥2 + 4𝑥 − 5 =(𝑥+2)

2√𝑥2 + 4𝑥 − 5 −

9

2log|(𝑥 + 2) + √𝑥2 + 4𝑥 − 5| +

𝐶

7. Integrate the function √1 + 3𝑥 − 𝑥2 [2

Marks]

Solution:

We need to integrate √1 + 3𝑥 − 𝑥2

Let 𝐼 = ∫ √1 + 3𝑥 − 𝑥2 𝑑𝑥

= ∫ √1 − (𝑥2 − 3𝑥 +9

4−

9

4) 𝑑𝑥

= ∫ √(1 +9

4) − (𝑥 −

3

2)

2

𝑑𝑥

= ∫ √(√13

2)

2

− (𝑥 −3

2)

2𝑑𝑥 [1

Mark]


2√𝑎2 − 𝑥2 +

𝑎2

2sin−1 𝑥

𝑎+ 𝐶

Here, 𝑎 =√13

2, 𝑥 = 𝑥 −

3

2

∴ 𝐼 =𝑥−

3

2

2 √1 + 3𝑥 − 𝑥2 +

13

4×2sin−1 (

𝑥−3

2

√13

2

) + 𝐶

=2𝑥−3

4√1 + 3𝑥 − 𝑥2 +

13

8sin−1 (

2𝑥−3

√13) + 𝐶

Hence, integration of √1 + 3𝑥 − 𝑥2 =2𝑥−3

4√1 + 3𝑥 − 𝑥2 +

13

8sin−1 (

2𝑥−3

√13) + 𝐶 [1

Mark]



8. Integrate the function √𝑥2 + 3𝑥 [2

Marks]

Solution:

We need to integrate √𝑥2 + 3𝑥

Let 𝐼 = ∫ √𝑥2 + 3𝑥 𝑑𝑥

= ∫ √𝑥2 + 3𝑥 +9

4−

9

4𝑑𝑥

= ∫ √(𝑥 +3

2)

2− (

3

2)

2𝑑𝑥 [1

Mark]

As we know that, ∫ √𝑥2 − 𝑎2 𝑑𝑥 =𝑥

2√𝑥2 − 𝑎2 −

𝑎2

2log|𝑥 + √𝑥2 − 𝑎2| + 𝐶

Here, 𝑎 =3

2, 𝑥 = 𝑥 +

3

2

∴ 𝐼 =(𝑥+

3

2)

2√𝑥2 + 3𝑥 −

9

4

2log |(𝑥 +

3

2) + √𝑥2 + 3𝑥| + 𝐶

=(2𝑥+3)

4√𝑥2 + 3𝑥 −

9

8log |(𝑥 +

3

2) + √𝑥2 + 3𝑥| + 𝐶

Hence, integration of √𝑥2 + 3𝑥 =(2𝑥+3)

4√𝑥2 + 3𝑥 −

9

8log |(𝑥 +

3

2) + √𝑥2 + 3𝑥| + 𝐶 [1

Mark]

9. Integrate the function √1 +𝑥2

9 [2 Marks]

Solution:

We need to integrate √1 +𝑥2

9

Let 𝐼 = ∫ √1 +𝑥2

9𝑑𝑥 =

1

3∫ √9 + 𝑥2 𝑑𝑥 =

1

3∫ √(3)2 + 𝑥2 𝑑𝑥 [1

Mark]

As we know that, ∫ √𝑥2 + 𝑎2 𝑑𝑥 =𝑥

2√𝑥2 + 𝑎2 +

𝑎2

2log|𝑥 + √𝑥2 + 𝑎2| + 𝐶



Here, 𝑎 = 3, 𝑥 = 𝑥

∴ 𝐼 =1

3[

𝑥

2√𝑥2 + 9 +

9

2log|𝑥 + √𝑥2 + 9|] + 𝐶

=𝑥

6√𝑥2 + 9 +

3

2log|𝑥 + √𝑥2 + 9| + 𝐶

Hence, integration of √1 +𝑥2

9=

𝑥

6√𝑥2 + 9 +

3

2log|𝑥 + √𝑥2 + 9| + 𝐶 [1

Mark]

10. ∫ √1 + 𝑥2 𝑑𝑥 is equal to [1

Mark]

(A) 𝑥

2√1 + 𝑥2 +

1

2log|𝑥 + √1 + 𝑥2| + 𝐶

(B) 2

3(1 + 𝑥2)

3

2 + 𝐶

(C) 2

3𝑥(1 + 𝑥2)

3

2 + 𝐶

(D) 𝑥2

2√1 + 𝑥2 +

1

2𝑥2 log|𝑥 + √1 + 𝑥2| + 𝐶

Solution:

We need to integrate ∫ √1 + 𝑥2 𝑑𝑥

As we know that, ∫ √𝑎2 + 𝑥2 𝑑𝑥 =𝑥

2√𝑎2 + 𝑥2 +

𝑎2

2log|𝑥 + √𝑥2 + 𝑎2| + 𝐶 [

𝟏

𝟐

Marks]

Here, 𝑎 = 1 and 𝑥 = 𝑥,

∴ ∫ √1 + 𝑥2 𝑑𝑥 =𝑥

2√1 + 𝑥2 +

1

2log|𝑥 + √1 + 𝑥2| + 𝐶

Hence, the correct Answer is (𝐴). [𝟏

𝟐

Marks]

11. ∫ √𝑥2 − 8𝑥 + 7 𝑑𝑥 is equal to [2 Marks]

(A) 1

2(𝑥 − 4)√𝑥2 − 8𝑥 + 7 + 9 log|𝑥 − 4 + √𝑥2 − 8𝑥 + 7| + 𝐶



(B) 1

2(𝑥 + 4)√𝑥2 − 8𝑥 + 7 + 9 log|𝑥 + 4 + √𝑥2 − 8𝑥 + 7| + 𝐶

(C) 1

2(𝑥 − 4)√𝑥2 − 8𝑥 + 7 − 3√2 log|𝑥 − 4 + √𝑥2 − 8𝑥 + 7| + 𝐶

(D) 1

2(𝑥 − 4)√𝑥2 − 8𝑥 + 7 −

9

2log|𝑥 − 4 + √𝑥2 − 8𝑥 + 7| + 𝐶

Solution:

We need to integrate ∫ √𝑥2 − 8𝑥 + 7 𝑑𝑥

Let 𝐼 = ∫ √𝑥2 − 8𝑥 + 7 𝑑𝑥

= ∫ √(𝑥2 − 8𝑥 + 16) − 9 𝑑𝑥

= ∫ √(𝑥 − 4)2 − (3)2 𝑑𝑥 [1

Mark]

As we know that, ∫ √𝑥2 − 𝑎2 𝑑𝑥 =𝑥

2√𝑥2 − 𝑎2 −

𝑎2

2log|𝑥 + √𝑥2 − 𝑎2| + 𝐶

Here, 𝑎 = 3, 𝑥 = 𝑥 − 4

∴ 𝐼 =(𝑥 − 4)

2√𝑥2 − 8𝑥 + 7 −

9

2log |(𝑥 − 4) + √𝑥2 − 8𝑥 + 7| + 𝐶

Hence, the correct answer is (𝐷) [1

Mark]

Exercise 7.8

1. Evaluate the definite integral ∫ 𝑥 𝑑𝑥𝑏

𝑎 as limit of sum. [4

Marks]

Solution:

We know that,

∫ 𝑓(𝑥) 𝑑𝑥 = (𝑏 − 𝑎) lim𝑛→∞

1

𝑛[𝑓(𝑎) + 𝑓(𝑎 + ℎ) + ⋯ + 𝑓(𝑎 + (𝑛 − 1)ℎ)],

𝑏

𝑎 where ℎ =

𝑏−𝑎

𝑛

Here we have, 𝑎 = 𝑎, 𝑏 = 𝑏 and 𝑓(𝑥) = 𝑥 [1

Mark]



∴ ∫ 𝑥𝑑𝑥 = (𝑏 − 𝑎)𝑏

𝑎lim

𝑛→∞

1

𝑛[𝑎 + (𝑎 + ℎ) … (𝑎 + 2ℎ) … 𝑎 + (𝑛 − 1)ℎ]

= (𝑏 − 𝑎) lim𝑛→∞

1

𝑛[(𝑎 + 𝑎 + 𝑎

𝑛 times + ⋯ + 𝑎) + (ℎ + 2ℎ + 3ℎ + ⋯ + (𝑛 − 1)ℎ)] [1

Mark]

= (𝑏 − 𝑎) lim𝑛→∞

1

𝑛[𝑛𝑎 + ℎ(1 + 2 + 3 + ⋯ + (𝑛 − 1))]

= (𝑏 − 𝑎) lim𝑛→∞

1

𝑛[𝑛𝑎 + ℎ {

(𝑛−1)(𝑛)

2}] [1

Mark]

= (𝑏 − 𝑎) lim𝑛→∞

1

𝑛[𝑛𝑎 +

𝑛(𝑛−1)ℎ

2]

= (𝑏 − 𝑎) lim𝑛→∞

𝑛

𝑛[𝑎 +

(𝑛−1)ℎ

2]

= (𝑏 − 𝑎) lim𝑛→∞

[𝑎 +(𝑛−1)ℎ

2]

= (𝑏 − 𝑎) lim𝑛→∞

[𝑎 +(𝑛−1)(𝑏−𝑎)

2𝑛]

= (𝑏 − 𝑎) lim𝑛→∞

[𝑎 +(1−

1

𝑛)(𝑏−𝑎)

2]

= (𝑏 − 𝑎) [𝑎 +(𝑏−𝑎)

2]

= (𝑏 − 𝑎) [2𝑎+𝑏−𝑎

2]

=(𝑏 − 𝑎)(𝑏 + 𝑎)

2

=1

2(𝑏2 − 𝑎2) [1

Mark]

2. Evaluate the definite integral ∫ (𝑥 + 1)𝑑𝑥5

0 as limit of sum. [4

Marks]

Solution:

Let 𝐼 = ∫ (𝑥 + 1)𝑑𝑥5

0

We know that,

∫ 𝑓(𝑥)𝑑𝑥𝑏

𝑎= (𝑏 − 𝑎) lim

𝑛→∞

1

𝑛[𝑓(𝑎) + 𝑓(𝑎 + ℎ) … 𝑓(𝑎 + (𝑛 − 1)ℎ)], where ℎ =

𝑏−𝑎

𝑛



Here we have, 𝑎 = 0, 𝑏 = 5 and 𝑓(𝑥) = (𝑥 + 1)

⇒ ℎ =5−0

𝑛=

5

𝑛 [1

Mark]

∴ ∫ (𝑥 + 1)𝑑𝑥 = (5 − 0) lim𝑛→∞

1

𝑛[𝑓(0) + 𝑓 (

5

𝑛) + ⋯ + 𝑓 ((𝑛 − 1)

5

𝑛)]

5

0

= 5 lim𝑛→∞

1

𝑛[1 + (

5

𝑛+ 1) +. . . {1 + (

5(𝑛−1)

𝑛)}]

= 5 lim𝑛→∞

1

𝑛[(1 + 1 + 1

𝑛 times … 1) + [

5

𝑛+ 2 ∙

5

𝑛+ 3 ∙

5

𝑛+ ⋯ (𝑛 − 1)

5

𝑛]] [1

Mark]

= 5 lim𝑛→∞

1

𝑛[𝑛 +

5

𝑛{1 + 2 + 3 … (𝑛 − 1)}]

= 5 lim𝑛→∞

1

𝑛[𝑛 +

5

𝑛∙

(𝑛−1)𝑛

2] [1

Mark]

= 5 lim𝑛→∞

1

𝑛[𝑛 +

5(𝑛−1)

2]

= 5 lim𝑛→∞

[1 +5

2(1 −

1

𝑛)]

= 5 [1 +5

2]

= 5 [7

2]

=35

2 [1

Mark]

3. Evaluate the definite integral ∫ 𝑥2𝑑𝑥3


Marks]

Solution:

We know that,

∫ 𝑓(𝑥)𝑑𝑥 = (𝑏 − 𝑎) lim𝑛→∞

1

𝑛

𝑏

𝑎

[𝑓(𝑎) + 𝑓(𝑎 + ℎ) + 𝑓(𝑎 + 2ℎ) … 𝑓{𝑎 + (𝑛 − 1)ℎ}], where ℎ =𝑏−𝑎

𝑛

Here we have, 𝑎 = 2, 𝑏 = 3 and 𝑓(𝑥) = 𝑥2



⇒ ℎ =3−2

𝑛=

1

𝑛 [1

Mark]

∴ ∫ 𝑥23

2𝑑𝑥 = (3 − 2) lim

𝑛→∞

1

𝑛[𝑓(2) + 𝑓 (2 +

1

𝑛) + 𝑓 (2 +

2

𝑛) … 𝑓 {2 + (𝑛 − 1)

1

𝑛}]

= lim𝑛→∞

1

𝑛[(2)2 + (2 +

1

𝑛)

2 + (2 +

2

𝑛)

2 + ⋯ (2 +

(𝑛−1)

𝑛)

2 ]

= lim𝑛→∞

1

𝑛[22 + {22 + (

1

𝑛)

2 + 2 · 2 ·

1

𝑛} + ⋯ + {(2)2 +

(𝑛−1)2

𝑛2 + 2 · 2 ·(𝑛−1)

𝑛}]

= lim𝑛→∞

1

𝑛[(22 + ⋯ +

𝑛 times22) + {(

1

𝑛)

2+ (

2

𝑛)

2+ ⋯ + (

𝑛−1

𝑛)

2} + 2 ∙ 2 ∙ {

1

𝑛+

2

𝑛+

3

𝑛+ ⋯ +

(𝑛−1)

𝑛}]

= lim𝑛→∞

1

𝑛[4𝑛 +

1

𝑛2{12 + 22 + 32 … +(𝑛 − 1)2} +

4

𝑛{1 + 2 + ⋯ + (𝑛 − 1)}] [1

Mark]

= lim𝑛→∞

1

𝑛[4𝑛 +

1

𝑛2 {𝑛(𝑛−1)(2𝑛−1)

6} +

4

𝑛{

𝑛(𝑛−1)

2}]

= lim𝑛→∞

1

𝑛[4𝑛 +

𝑛(1−1

𝑛)(2−

1

𝑛)

6+

4𝑛−4

2] [1

Mark]

= lim𝑛→∞

[4 +1

6(1 −

1

𝑛) (2 −

1

𝑛) + 2 −

2

𝑛]

= 4 +2

6+ 2

=19

3 [1

Mark]

4. Evaluate the definite integral ∫ (𝑥2 − 𝑥)𝑑𝑥4


Marks]

Solution:

Let 𝐼 = ∫ (𝑥2 − 𝑥)𝑑𝑥4

1

= ∫ 𝑥2𝑑𝑥 − ∫ 𝑥 𝑑𝑥

4

1

4

1

Let 𝐼 = 𝐼1 − 𝐼2 … (i)

where 𝐼1 = ∫ 𝑥2𝑑𝑥4

1 and 𝐼2 = ∫ 𝑥 𝑑𝑥

4

1



We know that,



𝑛→∞

1

𝑛[𝑓(𝑎) + 𝑓(𝑎 + ℎ) + 𝑓(𝑎 + (𝑛 − 1)ℎ)], where ℎ =

𝑏−𝑎

𝑛

Here, 𝐼1 = ∫ 𝑥2𝑑𝑥4

1,

𝑎 = 1, 𝑏 = 4 and 𝑓(𝑥) = 𝑥2

∴ ℎ =4−1

𝑛=

3

𝑛 [1

Mark]

𝐼1 = ∫ 𝑥24

1

𝑑𝑥 = (4 − 1) lim𝑛→∞

1

𝑛[𝑓(1) + 𝑓(1 + ℎ) + ⋯ + 𝑓(1 + (𝑛 − 1)ℎ)]

= 3 lim𝑛→∞

1

𝑛[12 + (1 +

3

𝑛)

2 + (1 + 2 ·

3

𝑛)

2 + ⋯ (1 +

(𝑛−1)3

𝑛)

2

]

= 3 lim𝑛→∞

1

𝑛[12 + {12 + (

3

𝑛)

2 + 2 ·

3

𝑛} + ⋯ + {12 + (

(𝑛−1)3

𝑛)

2

+2·(𝑛−1)·3

𝑛}]

= 3 lim𝑛→∞

1

𝑛[(12 + ⋯ +

𝑛 times12) + (

3

𝑛)

2{12 + 22 + ⋯ + (𝑛 − 1)2} + 2 ·

3

𝑛{1 + 2 + ⋯ + (𝑛 − 1)}] [1

Mark]

= 3 lim𝑛→∞

1

𝑛[𝑛 +

9

𝑛2 {(𝑛−1)(𝑛)(2𝑛−1)

6} +

6

𝑛{

(𝑛−1)(𝑛)

2}]

= 3 lim𝑛→∞

1

𝑛[𝑛 +

9𝑛

6(1 −

1

𝑛) (2 −

1

𝑛) +

6𝑛−6

2]

= 3 lim𝑛→∞

[1 +9

6(1 −

1

𝑛) (2 −

1

𝑛) + 3 −

3

𝑛]

= 3[1 + 3 + 3]

= 3[7]

Hence, 𝐼1 = 21 …(ii) [1

Mark]

Again, 𝐼2 = ∫ 𝑥𝑑𝑥4

1,

𝑎 = 1, 𝑏 = 4 and 𝑓(𝑥) = 𝑥

⇒ ℎ =4 − 1

𝑛=

3

𝑛

∴ 𝐼2 = (4 − 1) lim𝑛→∞

1

𝑛[𝑓(1) + 𝑓(1 + ℎ) + ⋯ 𝑓(𝑎 + (𝑛 − 1)ℎ)]

= 3 lim𝑛→∞

1

𝑛[1 + (1 + ℎ) + ⋯ + (1 + (𝑛 − 1)ℎ)]

= 3 lim𝑛→∞

1

𝑛[1 + (1 +

3

𝑛) + ⋯ + {1 + (𝑛 − 1)

3

𝑛}]



= 3 lim𝑛→∞

1

𝑛[(1 +1 + ⋯ +

𝑛 times1) +

3

𝑛(1 + 2 + ⋯ + (𝑛 − 1))] [1

Mark]

= 3 lim𝑛→∞

1

𝑛[𝑛 +

3

𝑛{

(𝑛 − 1)𝑛

2}]

= 3 lim𝑛→∞

1

𝑛[1 +

3

2(1 −

1

𝑛)]

= 3 [1 +3

2]

= 3 [5

2]

Hence, 𝐼2 =15

2 …(iii) [1

Mark]

From equations (i), (ii) and (iii), we get

𝐼 = 𝐼1 − 𝐼2 = 21 −15

2=

27

2 [1

Mark]

5. Evaluate the definite integral ∫ 𝑒𝑥𝑑𝑥1

−1 as limit of sum. [4 Marks]

Solution:

Let 𝐼 = ∫ 𝑒𝑥𝑑𝑥1

−1 …(i)

We know that,


1

𝑛[𝑓(𝑎) + 𝑓(𝑎 + ℎ) … 𝑓(𝑎 + (𝑛 − 1)ℎ)],

𝑏

𝑎 where ℎ =

𝑏−𝑎

𝑛

Here we have, 𝑎 = −1, 𝑏 = 1 and 𝑓(𝑥) = 𝑒𝑥

∴ ℎ =1+1

𝑛=

2

𝑛 [1

Mark]

∴ 𝐼 = (1 + 1) lim𝑛→∞

1

𝑛[𝑓(−1) + 𝑓 (−1 +

2

𝑛) + 𝑓 (−1 + 2 ·

2

𝑛) + ⋯ + 𝑓 (−1 +

(𝑛−1)2

𝑛)]

= 2 lim𝑛→∞

1

𝑛[𝑒−1 + 𝑒

(−1+2

𝑛)

+ 𝑒(−1+2∙

2

𝑛)

+ ⋯ 𝑒(−1+(𝑛−1)

2

𝑛)] [1

Mark]



= 2 lim𝑛→∞

1

𝑛[𝑒−1 {1 + 𝑒

2𝑛 + 𝑒

4𝑛 + 𝑒

6𝑛 + 𝑒

(𝑛−1)2𝑛}]

= 2 lim𝑛→∞

𝑒−1

𝑛[

𝑒2𝑛𝑛 −1

𝑒2𝑛−1

]

= 𝑒−1 × 2 lim𝑛→∞

1

𝑛[

𝑒2−1

𝑒2𝑛−1

] [1

Mark]

=𝑒−1×2(𝑒2−1)

lim2𝑛

→0(

𝑒2𝑛−1

2𝑛

)×2

= 𝑒−1 [2(𝑒2−1)

2] [lim

ℎ→0(

𝑒ℎ−1

ℎ) = 1]

=𝑒2−1

𝑒

= (𝑒 −1

𝑒) [1

Mark]

6. Evaluate the definite integral ∫ (𝑥 + 𝑒2𝑥)𝑑𝑥4


Marks]

Solution:

We know that,



𝑛→∞

1

𝑛[𝑓(𝑎) + 𝑓(𝑎 + ℎ) + ⋯ + 𝑓(𝑎 + (𝑛 − 1)ℎ)], where ℎ =

𝑏−𝑎

𝑛

Here we have, 𝑎 = 0, 𝑏 = 4 and 𝑓(𝑥) = 𝑥 + 𝑒2𝑥

∴ ℎ =4−0

𝑛=

4

𝑛 [1

Mark]

⇒ ∫ (𝑥 + 𝑒2𝑥)4

0

𝑑𝑥 = (4 − 0) lim𝑛→∞

1

𝑛[𝑓(0) + 𝑓(ℎ) + 𝑓(2ℎ) + ⋯ + 𝑓((𝑛 − 1)ℎ)]

= 4 lim𝑛→∞

1

𝑛[(0 + 𝑒0) + (ℎ + 𝑒2ℎ) + (2ℎ + 𝑒2∙2ℎ) + ⋯ + {(𝑛 − 1)ℎ + 𝑒2(𝑛−1)ℎ}]

= 4 lim𝑛→∞

1

𝑛[1 + (ℎ + 𝑒2ℎ) + (2ℎ + 𝑒4ℎ) + ⋯ + {(𝑛 − 1)ℎ + 𝑒2(𝑛−1)ℎ}]



= 4 lim𝑛→∞

1

𝑛[{ℎ + 2ℎ + 3ℎ + ⋯ + (𝑛 − 1)ℎ} + (1 + 𝑒2ℎ + 𝑒4ℎ + ⋯ + 𝑒2(𝑛−1)ℎ)] [2

Marks]

= 4 lim𝑛→∞

1

𝑛[ℎ{1 + 2 + ⋯ (𝑛 − 1)} + (

𝑒2ℎ𝑛−1

𝑒2ℎ−1)]

= 4 lim𝑛→∞

1

𝑛[

(ℎ(𝑛−1)𝑛)

2+ (

𝑒2ℎ𝑛−1

𝑒2ℏ−1)]

= 4 lim𝑛→∞

1

𝑛[

4

𝑛·

(𝑛−1)𝑛

2+ (

𝑒8−1

𝑒8𝑛−1

)]

= 4(2) + 4 lim𝑛→∞

(𝑒8−1)

(𝑒

8𝑛−1

8𝑛

)8

[2

Marks]

= 8 +4∙(𝑒8−1)

8(lim

𝑥→0

𝑒𝑥−1

𝑥= 1)

= 8 +𝑒8−1

2

=15+𝑒8

2 [1

Mark]

Exercise 7.9

1. Evaluate the definite integral ∫ (𝑥 + 1)𝑑𝑥1

−1 [2

Marks]

Solution:

Let 𝐼 = ∫ (𝑥 + 1)𝑑𝑥1

−1

∫(𝑥 + 1)𝑑𝑥 =𝑥2

2+ 𝑥 = 𝐹(𝑥) [1

Mark]

By second fundamental theorem of calculus, we get

𝐼 = 𝐹(1) − 𝐹(−1)

= (1

2+ 1) − (

1

2− 1)

=1

2+ 1 −

1

2+ 1



= 2 [1

Mark]

2. Evaluate the definite integral ∫1

𝑥𝑑𝑥

3

2

[2

Marks]

Solution:

Let 𝐼 = ∫1

𝑥𝑑𝑥

3

2

∫1

𝑥𝑑𝑥 = log|𝑥| = 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹(3) − 𝐹(2)

= log|3| − log|2| = log3

2 [1

Mark]

3. Evaluate the definite integral ∫ (4𝑥3 − 5𝑥2 + 6𝑥 + 9)𝑑𝑥2

1 [2

Marks]

Solution:

Let 𝐼 = ∫ (4𝑥3 − 5𝑥2 + 6𝑥 + 9)2

1𝑑𝑥

∫(4𝑥3 − 5𝑥2 + 6𝑥 + 9)𝑑𝑥 = 4 (𝑥4

4) − 5 (

𝑥3

3) + 6 (

𝑥2

2) + 9(𝑥)

= 𝑥4 −5𝑥3

3+ 3𝑥2 + 9𝑥 = 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹(2) − 𝐹(1)

𝐼 = {24 −5·(2)3

3+ 3(2)2 + 9(2)} − {(1)4 −

5(1)3

3+ 3(1)2 + 9(1)}



= (16 −40

3+ 12 + 18) − (1 −

5

3+ 3 + 9)

= 16 −40

3+ 12 + 18 − 1 +

5

3− 3 − 9

= 33 −35

3

=99−35

3

=64

3 [1

Mark]

4. Evaluate the definite integral ∫ sin 2𝑥 𝑑𝑥𝜋

40

[2

Marks]

Solution:

Let 𝐼 = ∫ sin 2𝑥 𝑑𝑥𝜋

40

∫ sin 2𝑥 𝑑𝑥 = (− cos 2𝑥

2) = 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹 (𝜋

4) − 𝐹(0)

= −1

2[cos 2 (

𝜋

4) − cos 0]

= −1

2[cos (

𝜋

2) − cos 0]

= −1

2[0 − 1]

=1

2 [1

Mark]

5. Evaluate the definite integral ∫ cos𝜋

20

2𝑥 𝑑𝑥 [2

Marks]

Solution:



Let 𝐼 = ∫ cos 2𝑥 𝑑𝑥𝜋

20

∫ cos 2𝑥 𝑑𝑥 = (sin 2𝑥

2) = 𝐹(𝑥) [1

Mark]


𝐼 = F (𝜋

2) − F(0)

=1

2[sin 2 (

𝜋

2) − sin 0]

=1

2[sin 𝜋 − sin 0]

=1

2[0 − 0] = 0 [1

Mark]

6. Evaluate the definite integral ∫ 𝑒𝑥𝑑𝑥5

4 [2 Marks]

Solution:

Let 𝐼 = ∫ 𝑒𝑥𝑑𝑥5

4

∫ 𝑒𝑥𝑑𝑥 = 𝑒𝑥 = 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹(5) − 𝐹(4)

= 𝑒5 − 𝑒4

= 𝑒4(𝑒 − 1) [1

Mark]

7. Evaluate the definite integral ∫ tan 𝑥 𝑑𝑥𝜋

40

[2

Marks]

Solution:

Let 𝐼 = ∫ tan 𝑥 𝑑𝑥𝜋

40



∫ tan 𝑥 𝑑𝑥 = − log|cos 𝑥| = 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹 (𝜋

4) − 𝐹(0)

= − log |cos𝜋

4| + log|cos 0|

= − log |1

√2| + log|1|

= − log(2)−1

2

=1

2log 2 [1

Mark]

8. Evaluate the definite integral ∫ cosec 𝑥 𝑑𝑥𝜋

4𝜋

6

[2

Marks]

Solution:

Let 𝐼 = ∫ cosec 𝑥 𝑑𝑥𝜋

4𝜋

6

∫ cosec 𝑥 𝑑𝑥 = log|cosec 𝑥 − cot 𝑥| = 𝐹(𝑥) [1

Mark]


𝐼 = F (𝜋

4) − F (

𝜋

6)

= log |cosec𝜋

4− cot

𝜋

4| − log |cosec

𝜋

6− cot

𝜋

6|

= log|√2 − 1| − log|2 − √3|

= log (√2−1

2−√3) [1

Mark]

9. Evaluate the definite integral ∫𝑑𝑥

√1−𝑥2

1

0 [2 Marks]



Solution:


√1−𝑥2

1

0

∫𝑑𝑥

√1−𝑥2= sin−1 𝑥 = 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹(1) − 𝐹(0)

= sin−1(1) − sin−1(0)

=𝜋

2− 0

=𝜋

2 [1

Mark]


1+𝑥2

1

0 [2 Marks]

Solution:


1+𝑥2

1

0

∫𝑑𝑥

1+𝑥2 = tan−1 𝑥 = 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹(1) − 𝐹(0)

= tan−1(1) − tan−1(0)

=𝜋

4 [1

Mark]


𝑥2−1

3

2 [2 Marks]



Solution:


𝑥2−1

3

2

∫𝑑𝑥

𝑥2−1=

1

2log |

𝑥−1

𝑥+1| = 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹(3) − 𝐹(2)

=1

2[log |

3−1

3+1| − log |

2−1

2+1|]

=1

2[log |

2

4| − log |

1

3|]

=1

2[log

1

2− log

1

3]

=1

2[log

3

2] [1

Mark]

12. Evaluate the definite integral ∫ cos2 𝑥 𝑑𝑥𝜋

20

[2

Marks]

Solution:

Let 𝐼 = ∫ cos2 𝑥 𝑑𝑥𝜋

20

∫ cos2 𝑥 𝑑𝑥 = ∫ (1+cos 2𝑥

2) 𝑑𝑥 =

𝑥

2+

sin 2𝑥

4=

1

2(𝑥 +

sin 2𝑥

2) = 𝐹(𝑥) [1

Mark]


𝐼 = [𝐹 (𝜋

2) − 𝐹(0)]

=1

2[(

𝜋

2+

sin 𝜋

2) − (0 +

sin 0

2)]

=1

2[

𝜋

2+ 0 − 0 − 0]

=𝜋

4 [1

Mark]



13. Evaluate the definite integral ∫𝑥𝑑𝑥

𝑥2+1

3

2 [2 Marks]

Solution:

Let 𝐼 = ∫𝑥

𝑥2+1

3

2𝑑𝑥

∫𝑥

𝑥2+1𝑑𝑥 =

1

2∫

2𝑥

𝑥2+1𝑑𝑥 =

1

2log(1 + 𝑥2) = 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹(3) − 𝐹(2)

=1

2[log(1 + (3)2) − log(1 + (2)2)]

=1

2[log(10) − log(5)]

=1

2log (

10

5) =

1

2log 2 [1

Mark]

14. Evaluate the definite integral ∫2𝑥+3

5𝑥2+1

1

0𝑑𝑥 [4

Marks]

Solution:

Let 𝐼 = ∫2𝑥+3

5𝑥2+1

1

0𝑑𝑥

∫2𝑥 + 3

5𝑥2 + 1𝑑𝑥 =

1

5∫

5(2𝑥 + 3)

5𝑥2 + 1𝑑𝑥

=1

5∫

10𝑥+15

5𝑥2+1𝑑𝑥

=1

5∫

10𝑥

5𝑥2+1𝑑𝑥 + 3 ∫

1

5𝑥2+1𝑑𝑥

=1

5∫

10𝑥

5𝑥2+1𝑑𝑥 + 3 ∫

1

5(𝑥2+1

5)

𝑑𝑥 [2

Marks]

=1

5log(5𝑥2 + 1) +

3

5∙

11

√5

tan−1 𝑥1

√5



=1

5log(5𝑥2 + 1) +

3

√5tan−1(√5𝑥)

= 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹(1) − 𝐹(0)

= {1

5log(5 + 1) +

3

√5tan−1(√5)} − {

1

5log(1) +

3

√5tan−1(0)}

=1

5log 6 +

3

√5tan−1 √5 [1

Mark]

15. Evaluate the definite integral ∫ 𝑥𝑒𝑥2𝑑𝑥

1

0 [2

Marks]

Solution:

Let 𝐼 = ∫ 𝑥𝑒𝑥2𝑑𝑥

1

0

Substitute 𝑥2 = 𝑡 ⇒ 2𝑥 𝑑𝑥 = 𝑑𝑡

When 𝑥 = 0, 𝑡 = 0 and when 𝑥 = 1, 𝑡 = 1,

∴ 𝐼 =1

2∫ 𝑒𝑡𝑑𝑡

1

0

1

2∫ 𝑒𝑡𝑑𝑡 =

1

2𝑒𝑡 = 𝐹(𝑡) [1

Mark]


𝐼 = 𝐹(1) − 𝐹(0)

=1

2𝑒 −

1

2𝑒0

=1

2(𝑒 − 1) [1

Mark]



16. Evaluate the definite integral ∫5𝑥2

𝑥2+4𝑥+3

2

1𝑑𝑥 [4

Marks]

Solution:

Let 𝐼 = ∫5𝑥2

𝑥2+4𝑥+3𝑑𝑥

2

1

Dividing 5𝑥2 by 𝑥2 + 4𝑥 + 3, we get

𝐼 = ∫ {5 −20𝑥+15

𝑥2+4𝑥+3} 𝑑𝑥

2

1

= ∫ 5𝑑𝑥 − ∫20𝑥+15

𝑥2+4𝑥+3

2

1𝑑𝑥

2

1

= [5𝑥]12 − ∫

20𝑥+15

𝑥2+4𝑥+3

2

1𝑑𝑥

= 5 × 2 − 5 − ∫20𝑥+15

𝑥2+4𝑥+3

2

1𝑑𝑥

= 5 − ∫20𝑥+15

𝑥2+4𝑥+3

2

1𝑑𝑥 [1

Mark]

Let 𝐼 = 5 − 𝐼1, where 𝐼1 = ∫20𝑥+15

𝑥2+4𝑥+3

2

1𝑑𝑥 …(i)

Now, consider 𝐼1 = ∫20𝑥+15

𝑥2+4𝑥+8

2

1𝑑𝑥

Let 20𝑥 + 15 = 𝐴𝑑

𝑑𝑥(𝑥2 + 4𝑥 + 3) + 𝐵

= 2𝐴𝑥 + (4𝐴 + 𝐵)


𝐴 = 10 and 𝐵 = −25

⇒ 𝐼1 = 10 ∫2𝑥+4

𝑥2+4𝑥+3

2

1𝑑𝑥 − 25 ∫

𝑑𝑥

𝑥2+4𝑥+3

2

1

Let 𝑥2 + 4𝑥 + 3 = 𝑡

⇒ (2𝑥 + 4)𝑑𝑥 = 𝑑𝑡 [1

Mark]

⇒ 𝐼1 = 10 ∫𝑑𝑡

𝑡− 25 ∫

𝑑𝑥

(𝑥 + 2)2 − 12

= 10 log 𝑡 − 25 [1

2log (

𝑥 + 2 − 1

𝑥 + 2 + 1)]

= [10 log(𝑥2 + 4𝑥 + 3)]12 − 25 [

1

2log (

𝑥 + 1

𝑥 + 3)]

1

2



= [10 log 15 − 10 log 8] − 25 [1

2log

3

5−

1

2log

2

4]

= [10 log(5 × 3) − 10 log(4 × 2)] −25

2[log 3 − log 5 − log 2 + log 4]

= [10 log 5 + 10 log 3 − 10 log 4 − 10 log 2] −25

2[log 3 − log 5 − log 2 + log 4]

= [10 +25

2] log 5 + [−10 −

25

2] log 4 + [10 −

25

2] log 3 + [−10 +

25

2] log 2

=45

2log 5 −

45

2log 4 −

5

2log 3 +

5

2log 2

=45

2log

5

4−

5

2log

3

2 [1

Mark]

Substituting the value of 𝐼1 in (i), we get

𝐼 = 5 − [45

2log

5

4−

5

2log

3

2]

= 5 −5

2[9 log

5

4− log

3

2] [1

Mark]

17. Evaluate the definite integral ∫ (2 sec2 𝑥 + 𝑥3 + 2)𝜋

40

𝑑𝑥 [2 Marks]

Solution:

Let 𝐼 = ∫ (2 sec2 𝑥 + 𝑥3 + 2)𝜋

40

𝑑𝑥

∫(2 sec2 𝑥 + 𝑥3 + 2)𝑑𝑥 = 2 tan 𝑥 +𝑥4

4+ 2𝑥 = 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹 (𝜋

4) − 𝐹(0)

= {(2 tan𝜋

4+

1

4(

𝜋

4)

4+ 2 (

𝜋

4)) − (2 tan 0 + 0 + 0)}

= 2 tan𝜋

4+

𝜋4

45 +𝜋

2

= 2 +𝜋

2+

𝜋4

1024 [1

Mark]



18. Evaluate the definite integral ∫ (sin2 𝑥

2− cos2 𝑥

2)

𝜋

0𝑑𝑥 [2 Marks]

Solution:

Let 𝐼 = ∫ (sin2 𝑥

2− cos2 𝑥

2)

𝜋

0𝑑𝑥

= − ∫ (cos2𝑥

2− sin2

𝑥

2)

𝜋

0

𝑑𝑥

= − ∫ cos 𝑥

𝜋

0

𝑑𝑥

− ∫ cos 𝑥 𝑑𝑥 = − sin 𝑥 = 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹(𝜋) − 𝐹(0)

= −sin 𝜋 + sin 0

= 0 [1

Mark]

19. Evaluate the definite integral ∫6𝑥+3

𝑥2+4

2

0𝑑𝑥 [4

Marks]

Solution:

Let 𝐼 = ∫6𝑥+3

𝑥2+4

2

0𝑑𝑥

∫6𝑥+3

𝑥2+4𝑑𝑥 = 3 ∫

2𝑥+1

𝑥2+4𝑑𝑥

= 3 ∫2𝑥

𝑥2+4𝑑𝑥 + 3 ∫

1

𝑥2+4𝑑𝑥 [1

Mark]

= 3 log(𝑥2 + 4) +3

2tan−1 𝑥

2= 𝐹(𝑥) [1

Mark]




𝐼 = 𝐹(2) − 𝐹(0)

= {3 log(22 + 4) +3

2tan−1 (

2

2)} − {3 log(0 + 4) +

3

2tan−1 (

0

2)}

= 3 log 8 +3

2tan−1 1 − 3 log 4 −

3

2tan−1 0

= 3 log 8 +3

2(

𝜋

4) − 3 log 4 − 0

= 3 log (8

4) +

3𝜋

8

= 3 log 2 +3𝜋

8 [2

Marks]

20. Evaluate the definite integral ∫ (𝑥𝑒𝑥 + sin𝜋𝑥

4) 𝑑𝑥

1

0 [2

Marks]

Solution:

Let 𝐼 = ∫ (𝑥𝑒𝑥 + sin𝜋𝑥

4) 𝑑𝑥

1

0

∫ (𝑥𝑒𝑥 + sin𝜋𝑥

4) 𝑑𝑥 = 𝑥 ∫ 𝑒𝑥𝑑𝑥 − ∫ {(

𝑑

𝑑𝑥𝑥) ∫ 𝑒𝑥𝑑𝑥} 𝑑𝑥 + {

− cos𝜋𝑥

4𝜋

4

}

= 𝑥𝑒𝑥 − ∫ 𝑒𝑥𝑑𝑥 −4

𝜋cos

𝜋𝑥

4

= 𝑥𝑒𝑥 − 𝑒𝑥 −4

𝜋cos

𝜋𝑥

4

= 𝐹(𝑥) [1

Mark]


𝐼 = 𝐹(1) − 𝐹(0)

= (1. 𝑒1 − 𝑒1 −4

𝜋cos

𝜋

4) − (0. 𝑒0 − 𝑒0 −

4

𝜋cos 0)

= 𝑒 − 𝑒 −4

𝜋(

1

√2) + 1 +

4

𝜋

= 1 +4

𝜋−

2√2

𝜋 [1

Mark]



21. ∫𝑑𝑥

1+𝑥2

√3

1 equals [2 Marks]

(A) 𝜋

3

(B) 2𝜋

3

(C) 𝜋

6

(D) 𝜋

12

Solution:

∫𝑑𝑥

1+𝑥2 = tan−1 𝑥 = 𝐹(𝑥) [1

Mark]


∫𝑑𝑥

1+𝑥2 = 𝐹(√3) − 𝐹(1)√3

1

= tan−1 √3 − tan−1 1

=𝜋

3−

𝜋

4

=𝜋

12

Hence, (𝐷) is the correct answer. [1

Mark]

22. ∫𝑑𝑥

4+9𝑥2

2

30

equals [2 Marks]

(A) 𝜋

6

(B) 𝜋

12

(C) 𝜋

24

(D) 𝜋

4

Solution:



∫𝑑𝑥

4+9𝑥2 = ∫𝑑𝑥

(2)2+(3𝑥)2

Put 3𝑥 = 𝑡 ⇒ 3𝑑𝑥 = 𝑑𝑡

∴ ∫𝑑𝑥

(2)2+(3𝑥)2 =1

3∫

𝑑𝑡

(2)2+𝑡2

=1

3[

1

2tan−1 𝑡

2]

=1

6tan−1 (

3𝑥

2)

= 𝐹(𝑥) [1

Mark]


∫𝑑𝑥

4+9𝑥2 = 𝐹 (2

3) − 𝐹(0)

2

30

=1

6tan−1 (

3

2∙

2

3) −

1

6tan−1 0

=1

6tan−1 1 − 0

=1

6×

𝜋

4

=𝜋

24

Hence, (𝐶) is the correct answer. [1

Mark]

Exercise 7.10

1. Evaluate the integral ∫𝑥

𝑥2+1

1

0𝑑𝑥 using substitution. [2

Marks]

Solution:

∫𝑥

𝑥2 + 1

1

0

𝑑𝑥

Let 𝑥2 + 1 = 𝑡 ⇒ 2𝑥 𝑑𝑥 = 𝑑𝑡

When 𝑥 = 0, 𝑡 = 1, when 𝑥 = 1, 𝑡 = 2



∴ ∫𝑥

𝑥2+1𝑑𝑥 =

1

2∫

𝑑𝑡

𝑡

2

1

1

0 [1

Mark]

=1

2[log|𝑡|]1

2

=1

2[log 2 − log 1]

=1

2log 2 [1

Mark]

2. Evaluate the integral ∫ √sin 𝜙 cos5 𝜙𝑑𝜙𝜋

20

using substitution. [2

Marks]

Solution:

Let 𝐼 = ∫ √sin 𝜙 cos5 𝜙𝑑𝜙𝜋

20

= ∫ √sin 𝜙 cos4 𝜙 cos 𝜙𝑑𝜙𝜋

20

Again, let sin 𝜙 = 𝑡 ⇒ cos 𝜙𝑑𝜙 = 𝑑𝑡

When 𝜙 = 0, 𝑡 = 0, when 𝜙 =𝜋

2, 𝑡 = 1

∴ 𝐼 = ∫ √𝑡1

0(1 − 𝑡2)2𝑑𝑡 [1

Mark]

= ∫ 𝑡1

21

0(1 + 𝑡4 − 2𝑡2)𝑑𝑡

= ∫ [𝑡1

2 + 𝑡9

2 − 2𝑡5

2] 𝑑𝑡1

0

= [𝑡

32

3

2

+𝑡

112

11

2

−2𝑡

72

7

2

]0

1

=2

3+

2

11−

4

7− 0

=154+42−132

231

=64

231 [1

Mark]



3. Evaluate the integral ∫ sin−1 (2𝑥

1+𝑥2) 𝑑𝑥1

0 using substitution. [4

Marks]

Solution:

Let 𝐼 = ∫ sin−1 (2𝑥

1+𝑥2) 𝑑𝑥1

0

Again, let 𝑥 = tan 𝜃 ⇒ 𝑑𝑥 = sec2 𝜃𝑑𝜃

When 𝑥 = 0, 𝜃 = 0, when 𝑥 = 1, 𝜃 =𝜋

4

𝐼 = ∫ sin−1 (2 tan 𝜃

1+tan2 𝜃) sec2 𝜃𝑑𝜃

𝜋

40

[1

Mark]

= ∫ sin−1(sin 2𝜃) sec2 𝜃𝑑𝜃𝜋

40

= ∫ 2𝜃 sec2 𝜃𝑑𝜃𝜋

40

= 2 ∫ 𝜃 sec2 𝜃𝑑𝜃𝜋

40

[1

Mark]

Taking 𝜃 as first function and sec2 𝜃 as second function and integrating using by parts, we get

𝐼 = 2 [𝜃 ∫ sec2 𝜃𝑑𝜃 − ∫ {(𝑑

𝑑𝑥𝜃) ∫ sec2 𝜃𝑑𝜃} 𝑑𝜃]

0

𝜋

4 [1

Mark]

= 2[𝜃 tan 𝜃 − ∫ tan 𝜃𝑑𝜃]0

𝜋

4

= 2[𝜃 tan 𝜃 + log|cos 𝜃|]0

𝜋4

= 2 [𝜋

4tan

𝜋

4+ log |cos

𝜋

4| − log|cos 0|]

= 2 [𝜋

4+ log (

1

√2) − log 1]

= 2 [𝜋

4−

1

2log 2]

=𝜋

2− log 2 [1

Mark]



4. Evaluate the integral ∫ 𝑥√𝑥 + 22

0 using substitution. (Put 𝑥 + 2 = 𝑡2) [2

Marks]

Solution:

∫ 𝑥√𝑥 + 2 𝑑𝑥2

0

Let 𝑥 + 2 = 𝑡2 ⇒ 𝑑𝑥 = 2𝑡𝑑𝑡

When 𝑥 = 0, 𝑡 = √2, when 𝑥 = 2, 𝑡 = 2

∴ ∫ 𝑥√𝑥 + 2 𝑑𝑥 = ∫ (𝑡2 − 2) √𝑡2 2𝑡𝑑𝑡2

√2

2

0 [1

Mark]

= 2 ∫ (𝑡4 − 2𝑡2) 𝑑𝑡2

√2

= 2 [𝑡5

5−

2𝑡3

3]

√2

2

= 2 [32

5−

16

3−

4√2

5+

4√2

3]

= 2 [96−80−12√2+20√2

15]

= 2 [16+8√2

15]

=16(2+√2)

15

=16√2(√2+1)

15 [1

Mark]

5. Evaluate the integral ∫sin 𝑥

1+cos2 𝑥

𝜋

20

𝑑𝑥 using substitution. [2

Marks]

Solution:

∫sin 𝑥

1+cos2 𝑥𝑑𝑥

𝜋

20

Let cos 𝑥 = 𝑡 ⇒ − sin 𝑥 𝑑𝑥 = 𝑑𝑡

When 𝑥 = 0, 𝑡 = 1, when 𝑥 =𝜋

2, 𝑡 = 0



⇒ ∫sin 𝑥

1+cos2 𝑥𝑑𝑥 = − ∫

𝑑𝑡

1+𝑡2

0

1

𝜋

20

[1

Mark]

= −[tan−1 𝑡]10

= −[tan−1 0 − tan−1 1]

= − [−𝜋

4]

=𝜋

4 [1

Mark]

6. Evaluate the integral ∫𝑑𝑥

𝑥+4−𝑥2

2


Marks]

Solution:

∫𝑑𝑥

𝑥+4−𝑥2 = ∫𝑑𝑥

−(𝑥2−𝑥−4)

2

0

2

0

= ∫𝑑𝑥

−(𝑥2−𝑥+1

4−

1

4−4)

2

0

= ∫𝑑𝑥

−[(𝑥−1

2)

2−

17

4]

2

0

= ∫𝑑𝑥

(√17

2)

2

−(𝑥−1

2)

2

2

0 [1

Mark]

Let 𝑥 −1

2= 𝑡 ⇒ 𝑑𝑥 = 𝑑𝑡

When 𝑥 = 0, 𝑡 = −1

2, when 𝑥 = 2, 𝑡 =

3

2

∴ ∫𝑑𝑥

(√17

2)

2

−(𝑥−1

2)

2

2

0= ∫

𝑑𝑡

(√17

2)

2

−𝑡2

3

2

−1

2

[1

Mark]

= [1

2(√17

2)

log√17

2+𝑡

√17

2−𝑡

]

−1

2

3

2

=1

√17[log

√17

2+

3

2

√17

2−

3

2

− log√17

2−

1

2

√17

2+

1

2

]



=1

√17[log

√17+3

√17−3− log

√17−1

√17+1]

=1

√17log

√17+3

√17−3×

√17+1

√17−1

=1

√17log [

17+3+4√17

17+3−4√17]

=1

√17log [

20+4√17

20−4√17]

=1

√17log (

5+√17

5−√17)

=1

√17log [

(5+√17)(5+√17)

25−17]

=1

√17log [

25+17+10√17

8]

=1

√17log (

42+10√17

8)

=1

√17log (

21+5√17

4) [2

Marks]

7. Evaluate the integral ∫𝑑𝑥

𝑥2+2𝑥+5

1

−1 using substitution. [4

Marks]

Solution:

∫𝑑𝑥

𝑥2+2𝑥+5= ∫

𝑑𝑥

(𝑥2+2𝑥+1)+4

1

−1= ∫

𝑑𝑥

(𝑥+1)2+(2)2

1

−1

1

−1 [1

Mark]

Let 𝑥 + 1 = 𝑡 ⇒ 𝑑𝑥 = 𝑑𝑡

When 𝑥 = −1, 𝑡 = 0, when 𝑥 = 1, 𝑡 = 2

∴ ∫𝑑𝑥

(𝑥+1)2+(2)2 = ∫𝑑𝑡

𝑡2+22

2

0

1

−1 [1

Mark]

= [1

2tan−1

𝑡

2]

0

2

=1

2tan−1 1 −

1

2tan−1 0



=1

2(

𝜋

4) =

𝜋

8 [2

Marks]

8. Evaluate the integral ∫ (1

𝑥−

1

2𝑥2) 𝑒2𝑥𝑑𝑥2


Marks]

Solution:

∫ (1

𝑥−

1

2𝑥2) 𝑒2𝑥𝑑𝑥2

1


When 𝑥 = 1, 𝑡 = 2, when 𝑥 = 2, 𝑡 = 4

∴ ∫ (1

𝑥−

1

2𝑥2) 𝑒2𝑥𝑑𝑥 =1

2∫ (

2

𝑡−

2

𝑡2) 𝑒𝑡𝑑𝑡4

2

2

1

= ∫ (1

𝑡−

1

𝑡2) 𝑒𝑡𝑑𝑡4

2 [2

Marks]

Let 1

𝑡= 𝑓(𝑡)

⇒ 𝑓′(𝑡) = −1

𝑡2

⇒ ∫ (1

𝑡−

1

𝑡2) 𝑒𝑡𝑑𝑡 = ∫ 𝑒𝑡[𝑓(𝑡) + 𝑓′(𝑡)]𝑑𝑡4

2

4

2

= [𝑒𝑡𝑓(𝑡)]24

= [𝑒𝑡 ∙1

𝑡]

2

4

= [𝑒𝑡

𝑡]

2

4

=𝑒4

4−

𝑒2

2

=𝑒2(𝑒2−2)

4 [2

Marks]

9. The value of the integral ∫(𝑥−𝑥3)

13

𝑥4 𝑑𝑥1

1

3

is [4 Marks]

(A) 6



(B) 0

(C) 3

(D) 4

Solution:

Let 𝐼 = ∫(𝑥−𝑥3)

13

𝑥4 𝑑𝑥1

1

3

= ∫𝑥 (

1

𝑥2−1)

13

𝑥4 𝑑𝑥1

1

3

= ∫(

1

𝑥2−1)

13

𝑥3 𝑑𝑥1

1

3

[1

Mark]

Again, let 1

𝑥2 − 1 = 𝑡 ⇒−2

𝑥3 𝑑𝑥 = 𝑑𝑡

⇒𝑑𝑥

𝑥3 = −1

2𝑑𝑡 [1

Mark]

When 𝑥 =1

3, 𝑡 = 8, when 𝑥 = 1, 𝑡 = 0

⇒ 𝐼 = −1

2∫ 𝑡

1

30

8𝑑𝑡 [1

Mark]

= −1

2[

𝑡43

4

3

]8

0

= −1

2[0 −

3

4× 16]

= 6

Hence, (𝐴) is the correct answer. [1

Mark]

10. If 𝑓(𝑥) = ∫ 𝑡 sin 𝑡 𝑑𝑡𝑥

0, then 𝑓′(𝑥) is [4

Marks]

(A) cos 𝑥 + 𝑥 sin 𝑥

(B) 𝑥 sin 𝑥

(C) 𝑥 cos 𝑥

(D) sin 𝑥 + 𝑥 cos 𝑥



Solution:

𝑓(𝑥) = ∫ 𝑡 sin 𝑡𝑑𝑡

𝑥

0

Using integration by parts, we get

𝑓(𝑥) = 𝑡 ∫ sin 𝑡 𝑑𝑡 − ∫ {(𝑑

𝑑𝑡𝑡) ∫ sin 𝑡 𝑑𝑡} 𝑑𝑡

𝑥

0

𝑥

0 [1

Mark]

= [𝑡(− cos 𝑡)]0𝑥 − ∫ (− cos 𝑡)

𝑥

0𝑑𝑡

= [−𝑡 cos 𝑡 + sin 𝑡]0𝑥 [1

Mark]

= −𝑥 cos 𝑥 + sin 𝑥

⇒ 𝑓′(𝑥) = −[{𝑥(−sin 𝑥)} + cos 𝑥] + cos 𝑥 [1

Mark]

= 𝑥 sin 𝑥 − cos 𝑥 + cos 𝑥

= 𝑥 sin 𝑥

Hence, (𝐵) is the correct answer. [1

Mark]

Exercise 7.11

1. By using the properties of definite integrals, evaluate the integral ∫ cos2 𝑥 𝑑𝑥𝜋

20

[2 Marks]

Solution:

Let 𝐼 = ∫ cos2 𝑥 𝑑𝑥𝜋

20

…(i)

⇒ 𝐼 = ∫ cos2 (𝜋

2− 𝑥) 𝑑𝑥

𝜋2

0

(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥) 𝑑𝑥

𝑎

0

𝑎

0

)

⇒ 𝐼 = ∫ sin2 𝑥 𝑑𝑥𝜋

20

…(ii) [1

Mark]

Adding (i) and (ii), we get



2𝐼 = ∫(sin2 𝑥 + cos2 𝑥)𝑑𝑥

𝜋2

0

⇒ 2𝐼 = ∫ 1 𝑑𝑥

𝜋2

0

⇒ 2𝐼 = [𝑥]0

𝜋2

⇒ 2𝐼 =𝜋

2

⇒ 𝐼 =𝜋

4 [1

Mark]

2. By using the properties of definite integrals, evaluate the integral ∫√sin 𝑥

√sin 𝑥+√cos 𝑥

𝜋

20

𝑑𝑥 [2

Marks]

Solution:

Let 𝐼 = ∫√sin 𝑥

√sin 𝑥+√cos 𝑥𝑑𝑥

𝜋

20

…(i)

⇒ 𝐼 = ∫√sin (

𝜋2

− 𝑥)

√sin (𝜋2 − 𝑥) + √cos (

𝜋2 − 𝑥)

𝑑𝑥

𝜋2

0

(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥

𝑎

0

𝑎

0

)

⇒ 𝐼 = ∫√cos 𝑥

√cos 𝑥 +√sin 𝑥𝑑𝑥

𝜋

20

…(ii) [1

Mark]


2𝐼 = ∫√sin 𝑥 + √cos 𝑥

√sin 𝑥 + √cos 𝑥

𝜋2

0

𝑑𝑥

⇒ 2𝐼 = ∫ 1 𝑑𝑥

𝜋2

0



⇒ 2𝐼 = [𝑥]0

𝜋2

⇒ 2𝐼 =𝜋

2

⇒ 𝐼 =𝜋

4 [1

Mark]

3. By using the properties of definite integrals, evaluate the integral ∫sin

32 𝑥𝑑𝑥

sin32 𝑥+cos

32 𝑥

𝜋

20

[2 Marks]

Solution:

Let 𝐼 = ∫sin

32 𝑥

sin32 𝑥+cos

32 𝑥

𝑑𝑥𝜋

20

…(i)

⇒ 𝐼 = ∫sin

32 (

𝜋2

− 𝑥)

sin32 (

𝜋2 − 𝑥) + cos

32 (

𝜋2 − 𝑥)

𝑑𝑥

𝜋2

0

(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥

𝑎

0

𝑎

0

)

⇒ 𝐼 = ∫cos

32 𝑥

sin32 𝑥+cos

32 𝑥

𝜋

20

𝑑𝑥 …(ii) [1

Mark]


2𝐼 = ∫sin

32 𝑥 + cos

32 𝑥

sin32 𝑥 + cos

32 𝑥

𝜋2

0

𝑑𝑥

⇒ 2𝐼 = ∫ 1 ∙ 𝑑𝑥

𝜋2

0

⇒ 2𝐼 = [𝑥]0

𝜋2

⇒ 2𝐼 =𝜋

2

⇒ 𝐼 =𝜋

4 [1

Mark]



4. By using the properties of definite integrals, evaluate the integral ∫cos5 𝑥𝑑𝑥

sin5 𝑥+cos5 𝑥

𝜋

20

[2 Marks]

Solution:

Let 𝐼 = ∫cos5 𝑥

sin5 𝑥+cos5 𝑥𝑑𝑥

𝜋

20

…(i)

⇒ 𝐼 = ∫cos5 (

𝜋2

− 𝑥)

sin5 (𝜋2

− 𝑥) + cos5 (𝜋2

− 𝑥)𝑑𝑥

𝜋2

0

(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥

𝑎

0

𝑎

0

)

⇒ 𝐼 = ∫sin5 𝑥

sin5 𝑥+cos5 𝑥

𝜋

20

𝑑𝑥 …(ii) [1

Mark]


2𝐼 = ∫sin5 𝑥 + cos5 𝑥

sin5 𝑥 + cos5 𝑥𝑑𝑥

𝜋2

0

⇒ 2𝐼 = ∫ 1 ∙ 𝑑𝑥

𝜋2

0

⇒ 2𝐼 = [𝑥]0

𝜋2

⇒ 2𝐼 =𝜋

2

⇒ 𝐼 =𝜋

4 [1

Mark]

5. By using the properties of definite integrals, evaluate the integral ∫ |𝑥 + 2|𝑑𝑥5

−5 [4 Marks]

Solution:

Let 𝐼 = ∫ |𝑥 + 2|𝑑𝑥5

−5

It is seen that (𝑥 + 2) ≤ 0 on [−5, −2] and (𝑥 + 2) ≥ 0 on [−2, 5].



∴ 𝐼 = ∫ −(𝑥 + 2)𝑑𝑥 + ∫ (𝑥 + 2)𝑑𝑥5

−2

−2

−5(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑓(𝑥)𝑑𝑥

𝑏

𝑐

𝑐

𝑎

𝑏

𝑎) [1

Mark]

𝐼 = − [𝑥2

2+ 2𝑥]

−5

−2

+ [𝑥2

2+ 2𝑥]

−2

5

[1

Mark]

= − [(−2)2

2+ 2(−2) −

(−5)2

2− 2(−5)] + [

(5)2

2+ 2(5) −

(−2)2

2− 2(−2)]

= − [2 − 4 −25

2+ 10] + [

25

2+ 10 − 2 + 4]

= −2 + 4 +25

2− 10 +

25

2+ 10 − 2 + 4

= 29 [2

Marks]

6. By using the properties of definite integrals, evaluate the integral ∫ |𝑥 − 5|𝑑𝑥8

2 [4 Marks]

Solution:

Let 𝐼 = ∫ |𝑥 − 5|𝑑𝑥8

2

It is seen that (𝑥 − 5) ≤ 0 on [2, 5] and (𝑥 − 5) ≥ 0 on [5, 8]. [1

Mark]

Hence, 𝐼 = ∫ −(𝑥 − 5)𝑑𝑥 + ∫ (𝑥 − 5)𝑑𝑥8

2

5

2(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑓(𝑥)𝑑𝑥

𝑏

𝑐

𝑐

𝑎

𝑏

𝑎) [1

Mark]

= − [𝑥2

2− 5𝑥]

2

5

+ [𝑥2

2− 5𝑥]

5

8

= − [25

2− 25 − 2 + 10] + [32 − 40 −

25

2+ 25]

= 9 [2

Marks]

7. By using the properties of definite integrals, evaluate the integral ∫ 𝑥(1 − 𝑥)𝑛1

0𝑑𝑥 [2

Marks]



Solution:

Let 𝐼 = ∫ 𝑥(1 − 𝑥)𝑛1

0𝑑𝑥

∴ 𝐼 = ∫ (1 − 𝑥)(1 − (1 − 𝑥))𝑛1

0𝑑𝑥 (∵ ∫ 𝑓(𝑥)

𝑎

0𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)

𝑎

0𝑑𝑥) [1

Mark]

= ∫ (1 − 𝑥)(𝑥)𝑛1

0𝑑𝑥

= ∫ (𝑥𝑛 − 𝑥𝑛+1)1

0𝑑𝑥

= [𝑥𝑛+1

𝑛+1−

𝑥𝑛+2

𝑛+2]

0

1

= [1

𝑛+1−

1

𝑛+2]

=(𝑛+2)−(𝑛+1)

(𝑛+1)(𝑛+2)

=1

(𝑛+1)(𝑛+2) [1

Mark]

8. By using the properties of definite integrals, evaluate the integral ∫ log(1 + tan 𝑥) 𝑑𝑥𝜋

40

[4

Marks]

Solution:

Let 𝐼 = ∫ log(1 + tan 𝑥) 𝑑𝑥𝜋

40

…(i)

∴ 𝐼 = ∫ log [1 + tan (𝜋

4− 𝑥)] 𝑑𝑥

𝜋

40

(∵ ∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥) 𝑑𝑥𝑎

0

𝑎

0) [1

Mark]

⇒ 𝐼 = ∫ log {1 +tan

𝜋

4−tan 𝑥

1+tan𝜋

4tan 𝑥

} 𝑑𝑥𝜋

40

⇒ 𝐼 = ∫ log {1 +1−tan 𝑥

1+tan 𝑥 } 𝑑𝑥

𝜋

40

⇒ 𝐼 = ∫ log2

(1+tan 𝑥)𝑑𝑥

𝜋

40

[1

Mark]

⇒ 𝐼 = ∫ log 2 𝑑𝑥𝜋

40

− ∫ log(1 + tan 𝑥) 𝑑𝑥𝜋

40



⇒ 𝐼 = ∫ log 2𝑑𝑥 − 𝐼𝜋

40

[From …(i)] [1

Mark]

⇒ 2𝐼 = [𝑥 log 2]0

𝜋4

⇒ 2𝐼 =𝜋

4log 2

⇒ 𝐼 =𝜋

8log 2 [1

Mark]

9. By using the properties of definite integrals, evaluate the integral ∫ 𝑥2

0√2 − 𝑥𝑑𝑥 [2

Marks]

Solution:

Let 𝐼 = ∫ 𝑥2

0√2 − 𝑥𝑑𝑥

𝐼 = ∫(2 − 𝑥)

2

0

√𝑥𝑑𝑥 (∵ ∫ 𝑓(𝑥)𝑑𝑥

𝑎

0

∫ 𝑓(𝑎 − 𝑥)

𝑎

0

𝑑𝑥)

= ∫ {2𝑥1

2 − 𝑥3

2}2

0𝑑𝑥

= [2 (𝑥

32

3

2

) −𝑥

52

5

2

]0

2

[1

Mark]

= [4

3𝑥

3

2 −2

5𝑥

5

2]0

2

=4

3(2)

3

2 −2

5(2)

5

2

=4×2√2

3−

2

5× 4√2

=8√2

3−

8√2

5

=40√2−24√2

15

=16√2

15 [1

Mark]



10. By using the properties of definite integrals, evaluate the integral ∫ (2 log sin 𝑥 − log sin 2𝑥)𝑑𝑥𝜋

20

[4

Marks]

Solution:

Let 𝐼 = ∫ (2 log sin 𝑥 − log sin 2𝑥)𝑑𝑥𝜋

20

⇒ 𝐼 = ∫ {2 log sin 𝑥 − log(2 sin 𝑥 cos 𝑥)}𝑑𝑥𝜋

20

⇒ 𝐼 = ∫ {2 log sin 𝑥 − log sin 𝑥 − log cos 𝑥 − log 2}𝑑𝑥𝜋

20

⇒ 𝐼 = ∫ {log sin 𝑥 − log cos 𝑥 − log 2}𝑑𝑥𝜋

20

…(i) [1

Mark]

As we know that, (∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥𝑎

0

𝑎

0)

⇒ 𝐼 = ∫ {log sin (𝜋

2− 𝑥) − log cos (

𝜋

2− 𝑥) − log 2} 𝑑𝑥

𝜋

20

⇒ 𝐼 = ∫ {log cos 𝑥 − log sin 𝑥 − log 2}𝑑𝑥𝜋

20

…(ii) [1

Mark]


2𝐼 = ∫(− log 2 − log 2)𝑑𝑥

𝜋2

0

⇒ 2𝐼 = −2 log 2 ∫ 1 ∙ 𝑑𝑥

𝜋2

0

⇒ 2𝐼 = −2 log 2 [𝑥]0

𝜋2

⇒ 𝐼 = − log 2 [𝜋

2]

⇒ 𝐼 =𝜋

2(− log 2)

⇒ 𝐼 =𝜋

2[log

1

2]



⇒ 𝐼 =𝜋

2log

1

2 [2

Marks]

11. By using the properties of definite integrals, evaluate the integral ∫ sin2 𝑥 𝑑𝑥𝜋

2−𝜋

2

[2 Marks]

Solution:

Let 𝐼 = ∫ sin2 𝑥𝑑𝑥𝜋

2−𝜋

2

Since, sin2(−𝑥) = (sin(−𝑥))2 = (−sin 𝑥)2 = sin2 𝑥, hence, sin2 𝑥 is an even function.

As we know that if 𝑓(𝑥) is an even function, then ∫ 𝑓(𝑥)𝑑𝑥 = 2 ∫ 𝑓(𝑥)𝑎

0𝑑𝑥

𝑎

−𝑎

Hence, 𝐼 = 2 ∫ sin2 𝑥𝑑𝑥𝜋

20

= 2 ∫1−cos 2𝑥

2𝑑𝑥

𝜋

20

[1

Mark]

= ∫ (1 − cos 2𝑥) 𝑑𝑥

𝜋2

0

= [𝑥 −sin 2𝑥

2]

0

𝜋2

=𝜋

2− 0 − 0 + 0

=𝜋

2 [1

Mark]

12. By using the properties of definite integrals, evaluate the integral ∫𝑥 𝑑𝑥

1+sin 𝑥

𝜋

0 [4

Marks]

Solution:

Let 𝐼 = ∫𝑥 𝑑𝑥

1+sin 𝑥

𝜋

0 …(i)



⇒ 𝐼 = ∫(𝜋 − 𝑥)

1 + sin(𝜋 − 𝑥)𝑑𝑥

𝜋

0

(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥

𝑎

0

𝑎

0

)

⇒ 𝐼 = ∫(𝜋−𝑥)

1+sin 𝑥𝑑𝑥

𝜋

0 …(ii) [2

Marks]


2𝐼 = ∫𝜋

1 + sin 𝑥𝑑𝑥

𝜋

0

⇒ 2𝐼 = 𝜋 ∫(1 − sin 𝑥)

(1 + sin 𝑥)(1 − sin 𝑥)𝑑𝑥

𝜋

0

⇒ 2𝐼 = 𝜋 ∫1 − sin 𝑥

cos2 𝑥𝑑𝑥

𝜋

0

⇒ 2𝐼 = 𝜋 ∫{sec2 𝑥 − tan 𝑥 sec 𝑥}𝑑𝑥

𝜋

0

⇒ 2𝐼 = 𝜋[tan 𝑥 − sec 𝑥]0𝜋 = 𝜋[tan 𝜋 − sec 𝜋 − tan 0 + sec 0]

⇒ 2𝐼 = 𝜋[2]

⇒ 𝐼 = 𝜋 [2

Marks]

13. By using the properties of definite integrals, evaluate the integral ∫ sin7 𝑥 𝑑𝑥𝜋

2

−𝜋

2

[2 Marks]

Solution:

Let 𝐼 = ∫ sin7 𝑥𝑑𝑥𝜋

2

−𝜋

2

…(i)

Since, sin7(−𝑥) = (sin(−𝑥))7 = (− sin 𝑥)7 = − sin7 𝑥, hence, sin2 𝑥 is an odd function.[1

Mark]

As we know that, if 𝑓(𝑥) is an odd function, then ∫ 𝑓(𝑥)𝑑𝑥 = 0𝑎

−𝑎



∴ 𝐼 = ∫ sin7 𝑥 𝑑𝑥 = 0𝜋

2

−𝜋

2

[1

Mark]

14. By using the properties of definite integrals, evaluate the integral ∫ cos5 𝑥 𝑑𝑥2𝜋

0 [4 Marks]

Solution:

Let 𝐼 = ∫ cos5 𝑥𝑑𝑥2𝜋

0 …(i)

cos5(2𝜋 − 𝑥) = cos5 𝑥 [1

Mark]

As we know that,

∫ 𝑓(𝑥)𝑑𝑥 = 2 ∫ 𝑓(𝑥)𝑑𝑥,𝑎

0

2𝜋

0 if 𝑓(2𝑎 − 𝑥) = 𝑓(𝑥) [1

Mark]

= 0 if 𝑓(2𝑎 − 𝑥) = −𝑓(𝑥)

∴ 𝐼 = 2 ∫ cos5 𝑥𝑑𝑥𝜋

0= 2 ∫ cos5 𝑥𝑑𝑥

2×𝜋

20

[1

Mark]

Again, using the same property, we have

⇒ 𝐼 = 2(0) = 0 [∵ cos5(𝜋 − 𝑥) = − cos5 𝑥] [1

Mark]

15. By using the properties of definite integrals, evaluate the integral ∫sin 𝑥−cos 𝑥

1+sin 𝑥 cos 𝑥𝑑𝑥

𝜋

20

[2

Marks]

Solution:

Let 𝐼 = ∫sin 𝑥−cos 𝑥


𝜋

20

…(i)

⇒ 𝐼 = ∫sin(

𝜋

2−𝑥)−cos(

𝜋

2−𝑥)

1+sin(𝜋

2−𝑥) cos(

𝜋

2−𝑥)

𝑑𝑥𝜋

20

(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥𝑎

0

𝑎

0)



⇒ 𝐼 = ∫cos 𝑥−sin 𝑥


𝜋

20

…(ii) [1

Mark]


2𝐼 = ∫0


𝜋

20

⇒ 𝐼 = 0 [1

Mark]

16. By using the properties of definite integrals, evaluate the integral ∫ log(1 + cos 𝑥) 𝑑𝑥𝜋

0

[6

Marks]

Solution:

Let 𝐼 = ∫ log(1 + cos 𝑥) 𝑑𝑥𝜋

0 …(i)

⇒ 𝐼 = ∫ log(1 + cos(𝜋 − 𝑥)) 𝑑𝑥

𝜋

0

(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)

𝑎

0

𝑑𝑥

𝑎

0

)

⇒ 𝐼 = ∫ log(1 − cos 𝑥) 𝑑𝑥𝜋

0 …(ii) [1

Mark]


2𝐼 = ∫ {log(1 + cos 𝑥) + log(1 − cos 𝑥)}𝑑𝑥𝜋

0

⇒ 2𝐼 = ∫ log(1 − cos2 𝑥) 𝑑𝑥𝜋

0

⇒ 2𝐼 = ∫ log sin2 𝑥 𝑑𝑥𝜋

0

⇒ 2𝐼 = 2 ∫ log sin 𝑥 𝑑𝑥𝜋

0

⇒ 𝐼 = ∫ log sin 𝑥 𝑑𝑥𝜋

0 …(iii) [1

Mark]

Since, sin(𝜋 − 𝑥) = sin 𝑥

∴ 𝐼 = 2 ∫ log sin 𝑥 𝑑𝑥𝜋

20

…(iv) [∫ 𝑓(𝑥)𝑑𝑥 = 2 ∫ 𝑓(𝑥)𝑑𝑥,𝑎

0

2𝜋

0 if 𝑓(2𝑎 − 𝑥) = 𝑓(𝑥)]



⇒ 𝐼 = 2 ∫ log sin (𝜋

2− 𝑥) 𝑑𝑥 = 2 ∫ log cos 𝑥 𝑑𝑥

𝜋

20

𝜋

20

…(v) [1

Mark]

Adding (iv) and (v), we get

2𝐼 = 2 ∫ (log sin 𝑥 + log cos 𝑥)𝑑𝑥𝜋

20

⇒ 𝐼 = ∫ (log sin 𝑥 + log cos 𝑥 + log 2 − log 2)𝑑𝑥𝜋

20

⇒ 𝐼 = ∫ (log 2 sin 𝑥 cos 𝑥 − log 2)𝑑𝑥𝜋

20

⇒ 𝐼 = ∫ log sin 2𝑥 𝑑𝑥 − ∫ log 2 𝑑𝑥𝜋

20

𝜋

20

⇒ 𝐼 = ∫ log sin 2𝑥 𝑑𝑥 −𝜋

20

log 2 [𝑥]0

𝜋

2

⇒ 𝐼 = ∫ log sin 2𝑥 𝑑𝑥 −𝜋

20

𝜋

2log 2 [2

Marks]



2, 𝑡 = 𝜋

∴ 𝐼 =1

2 ∫ log sin 𝑡 𝑑𝑡 −

𝜋

2log 2

𝜋

0

∴ 𝐼 =1

2 ∫ log sin 𝑥 𝑑𝑥 −

𝜋

2log 2

𝜋

0 [∵ ∫ 𝑓(𝑥)𝑑𝑥

𝑏

𝑎= ∫ 𝑓(𝑡)𝑑𝑡

𝑏

𝑎]

⇒ 𝐼 =1

2𝐼 −

𝜋

2log 2 [from (iii)]

⇒𝐼

2= −

𝜋

2log 2

⇒ 𝐼 = −𝜋 log 2 [1

Mark]

17. By using the properties of definite integrals, evaluate the integral ∫√𝑥

√𝑥+√𝑎−𝑥𝑑𝑥

𝑎

0 [2

Marks]

Solution:

Let 𝐼 = ∫√𝑥


𝑎

0 …(i)

As we know that, (∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥𝑎

0

𝑎

0)



Hence, 𝐼 = ∫√𝑎−𝑥

√𝑎−𝑥+√𝑥𝑑𝑥

𝑎

0 …(ii) [1

Mark]


2𝐼 = ∫√𝑥+√𝑎−𝑥


𝑎

0

⇒ 2𝐼 = ∫ 1 ∙ 𝑑𝑥𝑎

0

⇒ 2𝐼 = [𝑥]0𝑎

⇒ 2𝐼 = 𝑎

⇒ 𝐼 =𝑎

2 [1

Mark]

18. By using the properties of definite integrals, evaluate the integral ∫ |𝑥 − 1|𝑑𝑥4

0 [4 Marks]

Solution:

𝐼 = ∫ |𝑥 − 1|𝑑𝑥4

0

It is seen that, (𝑥 − 1) ≤ 0 when 0 ≤ 𝑥 ≤ 1 and (𝑥 − 1) ≥ 0 when 1 ≤ 𝑥 ≤ 4 [1

Mark]

Hence, 𝐼 = ∫ |𝑥 − 1|𝑑𝑥1

0+ ∫ |𝑥 − 1|𝑑𝑥

4

1(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑓(𝑥)

4

1

1

0

4

0𝑑𝑥) [1

Mark]

= ∫ −(𝑥 − 1)𝑑𝑥 + ∫ (𝑥 − 1)𝑑𝑥4

1

1

0

= [𝑥 −𝑥2

2]

0

1

+ [𝑥2

2− 𝑥]

1

4

= 1 −1

2+

(4)2

2− 4 −

1

2+ 1

= 1 −1

2+ 8 − 4 −

1

2+ 1

= 5 [2

Marks]



19. Show that ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 = 2 ∫ 𝑓(𝑥)𝑑𝑥,𝑎

0

𝑎

0 if 𝑓 and 𝑔 are defined as 𝑓(𝑥) = 𝑓(𝑎 − 𝑥) and

𝑔(𝑥) + 𝑔(𝑎 − 𝑥) = 4 [4

Marks]

Solution:

Let 𝐼 = ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥𝑎

0 …(i)

⇒ 𝐼 = ∫ 𝑓(𝑎 − 𝑥)𝑔(𝑎 − 𝑥)𝑑𝑥𝑎

0(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥

𝑎

0

𝑎

0)

⇒ 𝐼 = ∫ 𝑓(𝑥)𝑔(𝑎 − 𝑥)𝑑𝑥𝑎

0 …(ii) [2

Marks]


2𝐼 = ∫ {𝑓(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔(𝑎 − 𝑥)}𝑑𝑥𝑎

0

⇒ 2𝐼 = ∫ 𝑓(𝑥){𝑔(𝑥) + 𝑔(𝑎 − 𝑥)}𝑑𝑥𝑎

0

⇒ 2𝐼 = ∫ 𝑓(𝑥) × 4𝑑𝑥𝑎

0[∵ 𝑔(𝑥) + 𝑔(𝑎 − 𝑥) = 4]

⇒ 𝐼 = 2 ∫ 𝑓(𝑥)𝑑𝑥𝑎

0

Hence, ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 = 2 ∫ 𝑓(𝑥)𝑑𝑥𝑎

0

𝑎

0 [2

Marks]

20. The value of ∫ (𝑥3 + 𝑥 cos 𝑥 + tan5 𝑥 + 1)𝑑𝑥𝜋

2

−𝜋

2

is [4

Marks]

(A) 0

(B) 2

(C) 𝜋

(D) 1

Solution:

Let 𝐼 = ∫ (𝑥3 + 𝑥 cos 𝑥 + tan5 𝑥 + 1)𝑑𝑥𝜋

2

−𝜋

2



⇒ 𝐼 = ∫ 𝑥3𝑑𝑥 + ∫ 𝑥 cos 𝑥 + ∫ tan5 𝑥𝑑𝑥 + ∫ 1 ∙ 𝑑𝑥𝜋

2

−𝜋

2

𝜋

2

−𝜋

2

𝜋

2

−𝜋

2

𝜋

2

−𝜋

2

[1

Mark]

As we know that if 𝑓(𝑥) is an even function, then ∫ 𝑓(𝑥)𝑑𝑥 = 2 ∫ 𝑓(𝑥)𝑑𝑥𝑎

0

𝑎

−𝑎

And if 𝑓(𝑥) is an odd function, then ∫ 𝑓(𝑥)𝑑𝑥 = 0𝑎

−𝑎 [1

Mark]

Here, 𝑥3, 𝑥 cos 𝑥, and tan5 𝑥 are odd functions and 1 is an even function.

Hence, 𝐼 = 0 + 0 + 0 + 2 ∫ 1 ∙ 𝑑𝑥𝜋

20

= 2[𝑥]0

𝜋2

=2𝜋

2= 𝜋


Marks]

21. The value of ∫ log (4+3 sin 𝑥

4+3 cos 𝑥) 𝑑𝑥

𝜋

20

is [4

Marks]

(A) 2

(B) 3

4

(C) 0

(D) −2

Solution:

Let 𝐼 = ∫ log (4+3 sin 𝑥

4+3 cos 𝑥) 𝑑𝑥

𝜋

20

…(i)

⇒ 𝐼 = ∫ log [4+3 sin(

𝜋

2−𝑥)

4+3 cos(𝜋

2−𝑥)

] 𝑑𝑥𝜋

20

(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥𝑎

0

𝑎

0)

⇒ 𝐼 = ∫ log (4+3 cos 𝑥

4+3 sin 𝑥) 𝑑𝑥

𝜋

20

…(ii) [2

Marks]




2𝐼 = ∫ {log (4+3 sin 𝑥

4+3 cos 𝑥) + log (

4+3 cos 𝑥

4+3 sin 𝑥)}

𝜋

20

𝑑𝑥

⇒ 2𝐼 = ∫ log (4+3 sin 𝑥

4+3 cos 𝑥×

4+3 cos 𝑥

4+3 sin 𝑥) 𝑑𝑥

𝜋

20

⇒ 2𝐼 = ∫ log 1𝑑𝑥𝜋

20

⇒ 2𝐼 = ∫ 0 𝑑𝑥𝜋

20

⇒ 𝐼 = 0


Marks]

Miscellaneous Exercise on Chapter 7


𝑥−𝑥3 [4

Marks]

Solution:


𝑥−𝑥3

∫1

𝑥−𝑥3 𝑑𝑥 = ∫1

𝑥3(1

𝑥2−1)𝑑𝑥 [1

Mark]

Let 1

𝑥2 − 1 = 𝑡

⇒ −2


⇒𝑑𝑥

𝑥3 = −1

2𝑑𝑡 [1

Mark]

Hence, ∫1

𝑥−𝑥3 𝑑𝑥 = −1

2∫

1

𝑡𝑑𝑡

= −1

2log|𝑡| + 𝐶

= −1

2log |

1

𝑥2 − 1| + 𝐶

= −1

2log |

1−𝑥2

𝑥2 | + 𝐶



=1

2log |

𝑥2

1−𝑥2| + 𝐶 [2

Marks]


√𝑥+𝑎+√𝑥+𝑏 [4

Marks]

Solution:


√𝑥+𝑎+√𝑥+𝑏

1

√𝑥+𝑎+√𝑥+𝑏=

1

√𝑥+𝑎+√𝑥+𝑏×

√𝑥+𝑎−√𝑥+𝑏

√𝑥+𝑎−√𝑥+𝑏

=√𝑥+𝑎−√𝑥+𝑏

(𝑥+𝑎)−(𝑥+𝑏)

=(√𝑥+𝑎−√𝑥+𝑏)

𝑎−𝑏 [2

Marks]

⇒ ∫1

√𝑥+𝑎+√𝑥+𝑏𝑑𝑥 =

1

𝑎−𝑏∫(√𝑥 + 𝑎 − √𝑥 + 𝑏) 𝑑𝑥

=1

(𝑎−𝑏)[

(𝑥+𝑎)32

3

2

−(𝑥+𝑏)

32

3

2

] + 𝐶

=2

3(𝑎−𝑏)[(𝑥 + 𝑎)

3

2 − (𝑥 + 𝑏)3

2] + 𝐶 [2

Marks]


𝑥√𝑎𝑥−𝑥2[Hint: Put 𝑥 =

𝑎

𝑡] [4

Marks]

Solution:


𝑥√𝑎𝑥−𝑥2

Let 𝑥 =𝑎

𝑡⇒ 𝑑𝑥 = −

𝑎

𝑡2 𝑑𝑡



⇒ ∫1

𝑥√𝑎𝑥−𝑥2𝑑𝑥 = ∫

1

𝑎

𝑡√𝑎∙

𝑎

𝑡−(

𝑎

𝑡)

2(−

𝑎

𝑡2 𝑑𝑡) [1

Mark]

= − ∫1

𝑎𝑡∙

1

√1

𝑡−

1

𝑡2

𝑑𝑡

= −1

𝑎∫

1

√𝑡2

𝑡−

𝑡2

𝑡2

𝑑𝑡 [1

Mark]

= −1

𝑎∫

1

√𝑡−1𝑑𝑡

= −1

𝑎[2√𝑡 − 1] + 𝐶 [1

Mark]

= −1

𝑎[2√

𝑎

𝑥− 1] + 𝐶

= −2

𝑎√

𝑎−𝑥

𝑥+ 𝐶 [1

Mark]


𝑥2(𝑥4+1)34

[4 Marks]

Solution:


𝑥2(𝑥4+1)34

Multiplying and dividing by 𝑥−3, we get

𝑥−3

𝑥2∙𝑥−3(𝑥4+1)34

=𝑥−3(𝑥4+1)

−34

𝑥2∙𝑥−3 [1

Mark]

=(𝑥4+1)

−34

𝑥5∙(𝑥4)−34

=1

𝑥5 (𝑥4+1

𝑥4 )−

3

4



=1

𝑥5 (1 +1

𝑥4)−

3

4 [1

Mark]

Let 1

𝑥4 = 𝑡 ⇒ −4

𝑥5 𝑑𝑥 = 𝑑𝑡 ⇒1

𝑥5 𝑑𝑥 = −𝑑𝑡

4

∴ ∫1

𝑥2(𝑥4+1)34

𝑑𝑥 = ∫1

𝑥5 (1 +1

𝑥4)−

3

4𝑑𝑥 [1

Mark]

= −1

4∫(1 + 𝑡)−

34 𝑑𝑡

= −1

4[

(1+𝑡)14

1

4

] + 𝐶

= −1

4

(1+1

𝑥4)

14

1

4

+ 𝐶

= − (1 +1

𝑥4)

1

4+ 𝐶 [1

Mark]


𝑥12+𝑥

13

[Hint:1

𝑥12+𝑥

13

=1

𝑥13(1+𝑥

16)

, put 𝑥 = 𝑡6] [4

Marks]

Solution:


𝑥12+𝑥

13

=1

𝑥13(1+𝑥

16)

Let 𝑥 = 𝑡6 ⇒ 𝑑𝑥 = 6𝑡5𝑑𝑡

∴ ∫1

𝑥12+𝑥

13

𝑑𝑥 = ∫1

𝑥13(1+𝑥

16)

𝑑𝑥 [1

Mark]

= ∫6𝑡5

𝑡2(1+𝑡)𝑑𝑡

= 6 ∫𝑡3

(1+𝑡)𝑑𝑡 [1

Mark]



On dividing, we get

∫1

𝑥12+𝑥

13

𝑑𝑥 = 6 ∫ {(𝑡2 − 𝑡 + 1) −1

1+𝑡} 𝑑𝑡

= 6 [(𝑡3

3) − (

𝑡2

2) + 𝑡 − log|1 + 𝑡|]

= 2𝑥1

2 − 3𝑥1

3 + 6𝑥1

6 − 6 log (1 + 𝑥1

6) + 𝐶

= 2√𝑥 − 3𝑥1

3 + 6𝑥1

6 − 6 log (1 + 𝑥1

6) + 𝐶 [2

Marks]


(𝑥+1)(𝑥2+9) [4

Marks]

Solution:


(𝑥+1)(𝑥2+9)

Let 5𝑥

(𝑥+1)(𝑥2+9)=

𝐴

(𝑥+1)+

𝐵𝑥+𝐶

(𝑥2+9) …(i)

⇒ 5𝑥 = 𝐴(𝑥2 + 9) + (𝐵𝑥 + 𝐶)(𝑥 + 1)

⇒ 5𝑥 = 𝐴𝑥2 + 9𝐴 + 𝐵𝑥2 + 𝐵𝑥 + 𝐶𝑥 + 𝐶 [1

Mark]

On equating the coefficients of 𝑥2, 𝑥 and constant term, we get

𝐴 + 𝐵 = 0

𝐵 + 𝐶 = 5

9𝐴 + 𝐶 = 0

After solving these equations, we get

𝐴 = −1

2, 𝐵 =

1

2 and 𝐶 =

9

2 [1

Mark]

From equation (i), we get

5𝑥

(𝑥+1)(𝑥2+9)=

−1

2(𝑥+1)+

𝑥

2+

9

2

(𝑥2+9)

∫5𝑥

(𝑥+1)(𝑥2+9)𝑑𝑥 = ∫ {

−1

2(𝑥+1)+

(𝑥+9)

2(𝑥2+9)} 𝑑𝑥



We need to integrate 𝑒5 log 𝑥−𝑒4 log 𝑥

𝑒3 log 𝑥−𝑒2 log 𝑥

𝑒5 log 𝑥−𝑒4 log 𝑥

𝑒3 log 𝑥−𝑒2 log 𝑥 =𝑒4 log 𝑥(𝑒log 𝑥−1)

𝑒2 log 𝑥(𝑒log 𝑥−1) [1

Mark]

= 𝑒2 log 𝑥

= 𝑒log 𝑥2

= 𝑥2

∴ ∫𝑒5 log 𝑥−𝑒4 log 𝑥

𝑒3 log 𝑥−𝑒2 log 𝑥 𝑑𝑥 = ∫ 𝑥2𝑑𝑥 =𝑥3

3+ 𝐶 [1

Mark]

9. Integrate the function cos 𝑥

√4−sin2 𝑥 [2 Marks]

Solution:


√4−sin2 𝑥

Let sin 𝑥 = 𝑡 ⇒ cos 𝑥 𝑑𝑥 = 𝑑𝑡

⇒ ∫cos 𝑥

√4−sin2 𝑥𝑑𝑥 = ∫

𝑑𝑡

√(2)2−(𝑡)2 [1

Mark]

= sin−1 (𝑡

2) + 𝐶

= sin−1 (sin 𝑥

2) + 𝐶 [1

Mark]

10. Integrate the function sin8 𝑥−cos8 𝑥

1−2 sin2 𝑥 cos2 𝑥 [2 Marks]

Solution:

We need to integrate sin8 𝑥−cos8 𝑥

1−2 sin2 𝑥 cos2 𝑥



sin8 𝑥−cos8 𝑥

1−2 sin2 𝑥 cos2 𝑥=

(sin4 𝑥+cos4 𝑥)(sin4 𝑥−cos4 𝑥)

sin2 𝑥+cos2 𝑥−sin2 𝑥 cos2 𝑥−sin2 𝑥 cos2 𝑥 [1

Mark]

=(sin4 𝑥+cos4 𝑥)(sin2 𝑥+cos2 𝑥)(sin2 𝑥−cos2 𝑥)

(sin2 𝑥−sin2 𝑥 cos2 𝑥)+(cos2 𝑥−sin2 𝑥 cos2 𝑥)

=(sin4 𝑥+cos4 𝑥)(sin2 𝑥−cos2 𝑥)

sin2 𝑥(1−cos2 𝑥)+cos2 𝑥(1−sin2 𝑥)

=−(sin4 𝑥+cos4 𝑥)(cos2 𝑥−sin2 𝑥)

(sin4 𝑥+cos4 𝑥)

= − cos 2𝑥

∴ ∫sin8 𝑥−cos8 𝑥

1−2 sin2 𝑥 cos2 𝑥𝑑𝑥 = ∫ − cos 2𝑥 𝑑𝑥 = −

sin 2𝑥

2+ 𝐶 [1

Mark]


cos(𝑥+𝑎) cos(𝑥+𝑏) [4 Marks]

Solution:


cos(𝑥+𝑎) cos(𝑥+𝑏)

Multiplying and dividing by sin(𝑎 − 𝑏), we get

1

sin(𝑎−𝑏)[

sin(𝑎−𝑏)

cos(𝑥+𝑎) cos(𝑥+𝑏)]

=1

sin(𝑎−𝑏)[

sin[(𝑥+𝑎)−(𝑥+𝑏)]


=1

sin(𝑎−𝑏)[

sin(𝑥+𝑎) cos(𝑥+𝑏)−cos(𝑥+𝑎) sin(𝑥+𝑏)


=1

sin(𝑎−𝑏)[

sin(𝑥+𝑎)

cos(𝑥+𝑎)−

sin(𝑥+𝑏)

cos(𝑥+𝑏)]

=1

sin(𝑎−𝑏)[tan(𝑥 + 𝑎) − tan(𝑥 + 𝑏)] [2

Marks]

∫1

cos(𝑥+𝑎) cos(𝑥+𝑏)𝑑𝑥 =

1

sin(𝑎−𝑏)∫[tan(𝑥 + 𝑎) − tan(𝑥 + 𝑏)] 𝑑𝑥

=1

sin(𝑎−𝑏)[− log|cos(𝑥 + 𝑎)| + log|cos(𝑥 + 𝑏)|] + 𝐶

=1

sin(𝑎−𝑏)log |

cos(𝑥+𝑏)

cos(𝑥+𝑎)| + 𝐶 [2

Marks]




√1−𝑥8 [2

Marks]

Solution:

We need to integrate 𝑥3

√1−𝑥8

Let 𝑥4 = 𝑡 ⇒ 4𝑥3𝑑𝑥 = 𝑑𝑡

⇒ ∫𝑥3

√1−𝑥8𝑑𝑥 =

1

4∫

𝑑𝑡

√1−𝑡2 [1

Mark]

=1

4sin−1 𝑡 + 𝐶

=1

4sin−1(𝑥4) + 𝐶 [1

Mark]

13. Integrate the function 𝑒𝑥

(1+𝑒𝑥)(2+𝑒𝑥) [4

Marks]

Solution:

We need to integrate 𝑒𝑥

(1+𝑒𝑥)(2+𝑒𝑥)

Let 𝑒𝑥 = 𝑡 ⇒ 𝑒𝑥𝑑𝑥 = 𝑑𝑡

⇒ ∫𝑒𝑥

(1 + 𝑒𝑥)(2 + 𝑒𝑥)𝑑𝑥 = ∫

𝑑𝑡

(𝑡 + 1)(𝑡 + 2)

= ∫(𝑡 + 2) − (𝑡 + 1)

(𝑡 + 1)(𝑡 + 2)𝑑𝑡

= ∫ [1

(𝑡+1)−

1

(𝑡+2)] 𝑑𝑡 [2

Marks]

= log|𝑡 + 1| − log|𝑡 + 2| + 𝐶

= log |𝑡 + 1

𝑡 + 2| + 𝐶



= log |1+𝑒𝑥

2+𝑒𝑥| + 𝐶 [2

Marks]


(𝑥2+1)(𝑥2+4) [4

Marks]

Solution:


(𝑥2+1)(𝑥2+4)

∴1

(𝑥2+1)(𝑥2+4)=

𝐴𝑥+𝐵

(𝑥2+1)+

𝐶𝑥+𝐷

(𝑥2+4)

⇒ 1 = (𝐴𝑥 + 𝐵)(𝑥2 + 4) + (𝐶𝑥 + 𝐷)(𝑥2 + 1)

⇒ 1 = 𝐴𝑥3 + 4𝐴𝑥 + 𝐵𝑥2 + 4𝐵 + 𝐶𝑥3 + 𝐶𝑥 + 𝐷𝑥2 + 𝐷 [1

Mark]

On equating the coefficients of 𝑥3, 𝑥2, 𝑥 and constant term, we get

𝐴 + 𝐶 = 0

𝐵 + 𝐷 = 0

4𝐴 + 𝐶 = 0

4𝐵 + 𝐷 = 1 [1

Mark]


𝐴 = 0, 𝐵 =1

3, 𝐶 = 0 and 𝐷 = −

1

3 [1

Mark]


1

(𝑥2+1)(𝑥2+4)=

1

3(𝑥2+1)−

1

3(𝑥2+4)

∫1

(𝑥2+1)(𝑥2+4)𝑑𝑥 =

1

3∫

1

𝑥2+1𝑑𝑥 −

1

3∫

1

𝑥2+4𝑑𝑥

=1

3tan−1 𝑥 −

1

3∙

1

2tan−1 𝑥

2+ 𝐶

=1

3tan−1 𝑥 −

1

6tan−1 𝑥

2+ 𝐶 [1

Mark]



15. Integrate the function cos3𝑥 𝑒log sin 𝑥 [2 Marks]

Solution:

We need to integrate cos3𝑥 𝑒log sin 𝑥 = cos3𝑥 sin 𝑥

Let cos 𝑥 = 𝑡 ⇒ − sin 𝑥 𝑑𝑥 = 𝑑𝑡

⇒ ∫ cos3 𝑥𝑒log sin 𝑥𝑑𝑥 = ∫ cos3 𝑥 sin 𝑥𝑑𝑥 [1

Mark]

= − ∫ 𝑡3𝑑𝑡

= −𝑡4

4+ 𝐶

= −cos4 𝑥

4+ 𝐶 [1

Mark]

16. Integrate the function 𝑒3 log 𝑥(𝑥4 + 1)−1 [2

Marks]

Solution:

We need to integrate 𝑒3 log 𝑥(𝑥4 + 1)−1 = 𝑒log 𝑥3(𝑥4 + 1)−1 =

𝑥3

(𝑥4+1)

Let 𝑥4 + 1 = 𝑡 ⇒ 4𝑥3𝑑𝑥 = 𝑑𝑡

⇒ ∫ 𝑒3 log 𝑥(𝑥4 + 1)−1𝑑𝑥 = ∫𝑥3

(𝑥4 + 1)𝑑𝑥

=1

4∫

𝑑𝑡

𝑡 [1

Mark]

=1

4log|𝑡| + 𝐶

=1

4log|𝑥4 + 1| + 𝐶

=1

4log(𝑥4 + 1) + 𝐶 [1

Mark]



17. Integrate the function 𝑓′(𝑎𝑥 + 𝑏)[𝑓(𝑎𝑥 + 𝑏)]𝑛 [2 Marks]

Solution:

We need to integrate 𝑓′(𝑎𝑥 + 𝑏)[𝑓(𝑎𝑥 + 𝑏)]𝑛

Let 𝑓(𝑎𝑥 + 𝑏) = 𝑡 ⇒ 𝑎𝑓′(𝑎𝑥 + 𝑏)𝑑𝑥 = 𝑑𝑡

⇒ ∫ 𝑓′(𝑎𝑥 + 𝑏) [𝑓(𝑎𝑥 + 𝑏)]𝑛𝑑𝑥 =1

𝑎∫ 𝑡𝑛𝑑𝑡 [1

Mark]

=1

𝑎[

𝑡𝑛+1

𝑛 + 1] + 𝐶

=1

𝑎(𝑛+1)(𝑓(𝑎𝑥 + 𝑏))

𝑛+1+ 𝐶 [1

Mark]


√sin3 𝑥 sin(𝑥+𝛼) [4 Marks]

Solution:


√sin3 𝑥 sin(𝑥+𝛼)

1

√sin3 𝑥 sin(𝑥+𝛼)=

1

√sin3 𝑥(sin 𝑥 cos 𝛼+cos 𝑥 sin 𝛼)

=1

√sin4 𝑥 cos 𝛼+sin3 𝑥 cos 𝑥 sin 𝛼

=1

sin2 𝑥√cos 𝛼+cot 𝑥 sin 𝛼

=cosec2 𝑥

√cos 𝛼+cot 𝑥 sin 𝛼 [2

Marks]

Let cos 𝛼 + cot 𝑥 sin 𝛼 = 𝑡 ⇒ − cosec2 𝑥 sin 𝛼 𝑑𝑥 = 𝑑𝑡

∴ ∫1

sin3 𝑥 sin(𝑥+𝛼)𝑑𝑥 = ∫

cosec2 𝑥

√cos 𝛼+cot 𝑥 sin 𝛼𝑑𝑥

=−1

sin 𝛼∫

𝑑𝑡

√𝑡

=−1

sin 𝛼[2√𝑡] + 𝐶



=−1

sin 𝛼[2√cos 𝛼 + cot 𝑥 sin 𝛼] + 𝐶

=−2

sin 𝛼√cos 𝛼 +

cos 𝑥 sin 𝛼

sin 𝑥+ 𝐶

=−2

sin 𝛼√

sin 𝑥 cos 𝛼+cos 𝑥 sin 𝛼

sin 𝑥+ 𝐶

= −2

sin 𝛼√

sin(𝑥+𝛼)

sin 𝑥+ 𝐶 [2

Marks]

19. Integrate the function sin−1 √𝑥−cos−1 √𝑥

sin−1 √𝑥+cos−1 √𝑥, 𝑥 ∈ [0, 1] [6

Marks]

Solution:

We need to integrate sin−1 √𝑥−cos−1 √𝑥

sin−1 √𝑥+cos−1 √𝑥, 𝑥 ∈ [0, 1]

Let 𝐼 = ∫sin−1 √𝑥−cos−1 √𝑥

sin−1 √𝑥+cos−1 √𝑥𝑑𝑥

As we know that, sin−1𝑥 + cos−1𝑥 =𝜋

2

Hence, sin−1 √𝑥 + cos−1 √𝑥 =𝜋

2 [1

Mark]

⇒ 𝐼 = ∫(

𝜋

2−cos−1 √𝑥)−cos−1 √𝑥

𝜋

2

𝑑𝑥

=2

𝜋∫ (

𝜋

2− 2 cos−1 √𝑥) 𝑑𝑥

=2

𝜋∙

𝜋

2∫ 1 ∙ 𝑑𝑥 −

4

𝜋∫ cos−1 √𝑥 𝑑𝑥

= 𝑥 −4

𝜋∫ cos−1 √𝑥 𝑑𝑥 …(i) [1

Mark]

Let 𝐼1 = ∫ cos−1 √𝑥 𝑑𝑥

Also, consider 𝑥 = 𝑡2 ⇒ 𝑑𝑥 = 2𝑡 𝑑𝑡

⇒ 𝐼1 = 2 ∫ cos−1 𝑡 ∙ 𝑡 𝑑𝑡



= 2 [cos−1 𝑡 ∙𝑡2

2− ∫

−1

√1−𝑡2∙

𝑡2

2𝑑𝑡] (using integration by parts) [1

Mark]

= 𝑡2 cos−1 𝑡 + ∫𝑡2


= 𝑡2 cos−1 𝑡 − ∫1−𝑡2−1


= 𝑡2 cos−1 𝑡 − ∫ √1 − 𝑡2𝑑𝑡 + ∫1


= 𝑡2 cos−1 𝑡 −𝑡

2√1 − 𝑡2 −

1

2sin−1 𝑡 + sin−1 𝑡

= 𝑡2 cos−1 𝑡 −𝑡

2√1 − 𝑡2 +

1

2sin−1 𝑡

= 𝑥 cos−1 √𝑥 −√𝑥

2√1 − 𝑥 +

1

2sin−1 √𝑥 [1

Mark]


𝐼 = 𝑥 −4

𝜋[𝑥 cos−1 √𝑥 −

√𝑥

2√1 − 𝑥 +

1

2sin−1 √𝑥]

= 𝑥 −4

𝜋[𝑥 (

𝜋

2− sin−1 √𝑥) −

√𝑥−𝑥2

2+

1

2sin−1 √𝑥]

= 𝑥 − 2𝑥 +4𝑥

𝜋sin−1 √𝑥 +

2

𝜋√𝑥 − 𝑥2 −

2

𝜋sin−1 √𝑥

= −𝑥 +2

𝜋[(2𝑥 − 1) sin−1 √𝑥] +

2

𝜋√𝑥 − 𝑥2 + 𝐶

=2(2𝑥−1)

𝜋sin−1 √𝑥 +

2

𝜋√𝑥 − 𝑥2 − 𝑥 + 𝐶 [2

Marks]

20. Integrate the function √1−√𝑥

1+√𝑥 [6 Marks]

Solution:

We need to integrate √1−√𝑥

1+√𝑥

Let 𝐼 = ∫ √1−√𝑥

1+√𝑥𝑑𝑥

Again, let 𝑥 = cos2 𝜃 ⇒ 𝑑𝑥 = −2 sin 𝜃 cos 𝜃𝑑𝜃 [1

Mark]



𝐼 = ∫ √1−cos 𝜃

1+cos 𝜃(−2 sin 𝜃 cos 𝜃)𝑑𝜃

= − ∫ √2 sin2𝜃

2

2 cos2𝜃

2

2 sin 𝜃 cos 𝜃 𝑑𝜃

= −2 ∫ tan𝜃

2∙ sin 𝜃 cos 𝜃𝑑𝜃

= −2 ∫sin

𝜃

2

cos𝜃

2

(2 sin𝜃

2cos

𝜃

2) cos 𝜃𝑑𝜃

= −4 ∫ sin2 𝜃

2cos 𝜃𝑑𝜃

= −4 ∫ sin2 𝜃

2∙ (2 cos2 𝜃

2− 1) 𝑑𝜃

= −4 ∫ (2 sin2 𝜃

2cos2 𝜃

2− sin2 𝜃

2) 𝑑𝜃

= −8 ∫ sin2 𝜃

2∙ cos2 𝜃

2𝑑𝜃 + 4 ∫ sin2 𝜃

2𝑑𝜃

= −2 ∫ sin2 𝜃𝑑𝜃 + 4 ∫ sin2 𝜃

2𝑑𝜃

= −2 ∫ (1−cos 2𝜃

2) 𝑑𝜃 + 4 ∫

1−cos 𝜃

2𝑑𝜃 [2

Marks]

= −2 [𝜃

2−

sin 2𝜃

4] + 4 [

𝜃

2−

sin 𝜃

2] + 𝐶

= −𝜃 +sin 2𝜃

2+ 2𝜃 − 2 sin 𝜃 + 𝐶

= 𝜃 +sin 2𝜃

2− 2 sin 𝜃 + 𝐶

= 𝜃 +2 sin 𝜃 cos 𝜃

2− 2 sin 𝜃 + 𝐶

= 𝜃 + √1 − cos2 𝜃 ∙ cos 𝜃 − 2√1 − cos2 𝜃 + 𝐶

= cos−1 √𝑥 + √1 − 𝑥 ∙ √𝑥 − 2√1 − 𝑥 + 𝐶

= −2√1 − 𝑥 + cos−1 √𝑥 + √𝑥(1 − 𝑥) + 𝐶

= −2√1 − 𝑥 + cos−1 √𝑥 + √𝑥 − 𝑥2 + 𝐶 [3

Marks]

21. Integrate the function 2+sin 2𝑥

1+cos 2𝑥𝑒𝑥 [4

Marks]



Solution:

We need to integrate 2+sin 2𝑥

1+cos 2𝑥𝑒𝑥

Let, 𝐼 = ∫ (2+sin 2𝑥

1+cos 2𝑥) 𝑒𝑥 𝑑𝑥

= ∫ (2+2 sin 𝑥 cos 𝑥

2 cos2 𝑥) 𝑒𝑥

= ∫ (1+sin 𝑥 cos 𝑥

cos2 𝑥) 𝑒𝑥

= ∫(sec2 𝑥 + tan 𝑥)𝑒𝑥 [2

Marks]

Let 𝑓(𝑥) = tan 𝑥 ⇒ 𝑓′(𝑥) = sec2 𝑥

∴ 𝐼 = ∫(𝑓(𝑥) + 𝑓′(𝑥)]𝑒𝑥 𝑑𝑥

= 𝑒𝑥𝑓(𝑥) + 𝐶

= 𝑒𝑥 tan 𝑥 + 𝐶 [2

Marks]

22. Integrate the function 𝑥2+𝑥+1

(𝑥+1)2(𝑥+2) [4

Marks]

Solution:

We need to integrate 𝑥2+𝑥+1

(𝑥+1)2(𝑥+2)

Let 𝑥2+𝑥+1

(𝑥+1)2(𝑥+2)=

𝐴

(𝑥+1)+

𝐵

(𝑥+1)2 +𝐶

(𝑥+2) …(i) [1

Mark]

⇒ 𝑥2 + 𝑥 + 1 = 𝐴(𝑥 + 1)(𝑥 + 2) + 𝐵(𝑥 + 2) + 𝐶(𝑥2 + 2𝑥 + 1)

⇒ 𝑥2 + 𝑥 + 1 = 𝐴(𝑥2 + 3𝑥 + 2) + 𝐵(𝑥 + 2) + 𝐶(𝑥2 + 2𝑥 + 1)

⇒ 𝑥2 + 𝑥 + 1 = (𝐴 + 𝐶)𝑥2 + (3𝐴 + 𝐵 + 2𝐶)𝑥 + (2𝐴 + 2𝐵 + 𝐶) [1 Mark]

After equating the coefficients of 𝑥2, 𝑥 and constant term, we get

𝐴 + 𝐶 = 1

3𝐴 + 𝐵 + 2𝐶 = 1



2𝐴 + 2𝐵 + 𝐶 = 1


𝐴 = −2, 𝐵 = 1 and 𝐶 = 3 [1 Mark]


𝑥2+𝑥+1

(𝑥+1)2(𝑥+2)=

−2

(𝑥+1)+

3

(𝑥+2)+

1

(𝑥+1)2

Hence, ∫𝑥2+𝑥+1

(𝑥+1)2(𝑥+2)𝑑𝑥 = −2 ∫

1

𝑥+1𝑑𝑥 + 3 ∫

1

(𝑥+2)𝑑𝑥 + ∫

1

(𝑥+1)2 𝑑𝑥

= −2 log|𝑥 + 1| + 3 log|𝑥 + 2| −1

𝑥+1+ 𝐶 [1

Mark]

23. Integrate the function tan−1 √1−𝑥

1+𝑥 [4

Marks]

Solution:

We need to integrate tan−1 √1−𝑥

1+𝑥

Let 𝐼 = ∫ tan−1 √1−𝑥

1+𝑥𝑑𝑥

Again, let 𝑥 = cos 𝜃 ⇒ 𝑑𝑥 = − sin 𝜃𝑑𝜃 [1 Mark]

𝐼 = ∫ tan−1 √1−cos 𝜃

1+cos 𝜃(− sin 𝜃𝑑𝜃)

= − ∫ tan−1 √2 sin2𝜃

2

2 cos2𝜃

2

sin 𝜃𝑑𝜃

= − ∫ tan−1 tan𝜃

2∙ sin 𝜃𝑑𝜃

= −1

2∫ 𝜃 ∙ sin 𝜃𝑑𝜃 [1

Mark]

= −1

2[𝜃 ∙ (− cos 𝜃) − ∫ 1 ∙ (− cos 𝜃)𝑑𝜃] (using integration by parts)

= −1

2[−𝜃 cos 𝜃 + sin 𝜃] + 𝐶



=1

2𝜃 cos 𝜃 −

1

2sin 𝜃 + 𝐶

=1

2cos−1 𝑥 ∙ 𝑥 −

1

2√1 − 𝑥2 + 𝐶

=𝑥

2cos−1 𝑥 −

1

2√1 − 𝑥2 + 𝐶

=1

2(𝑥 cos−1 𝑥 − √1 − 𝑥2) + 𝐶 [2

Marks]

24. Integrate the function √𝑥2+1[log(𝑥2+1)−2 log 𝑥]

𝑥4 [6 Marks]

Solution:

We need to integrate √𝑥2+1[log(𝑥2+1)−2 log 𝑥]

𝑥4

√𝑥2+1[log(𝑥2+1)−2 log 𝑥]

𝑥4 =√𝑥2+1

𝑥4[log(𝑥2 + 1) − log 𝑥2]

=√𝑥2+1

𝑥4 [log (𝑥2+1

𝑥2 )]

=√𝑥2+1

𝑥4 log (1 +1

𝑥2)

=1

𝑥3√

𝑥2+1

𝑥2 log (1 +1

𝑥2)

=1

𝑥3√1 +

1

𝑥2 log (1 +1

𝑥2) [2

Marks]

Let 1 +1

𝑥2 = 𝑡 ⇒−2


∴ 𝐼 = ∫√𝑥2+1[log(𝑥2+1)−2 log 𝑥]

𝑥4 𝑑𝑥 [1

Mark]

= ∫1

𝑥3 √1 +1

𝑥2 log (1 +1

𝑥2) 𝑑𝑥

= −1

2∫ √𝑡 log 𝑡 𝑑𝑡

= −1

2∫ 𝑡

1

2 ∙ log 𝑡 𝑑𝑡 [1

Mark]




𝐼 = −1

2[log 𝑡 ∙ ∫ 𝑡

1

2𝑑𝑡 − ∫ {(𝑑

𝑑𝑡log 𝑡) ∫ 𝑡

1

2𝑑𝑡} 𝑑𝑡]

= −1

2[log 𝑡 ∙

𝑡32

3

2

− ∫1

𝑡∙

𝑡32

3

2

𝑑𝑡]

= −1

2[

2

3𝑡

3

2 log 𝑡 −2

3∫ 𝑡

1

2𝑑𝑡]

= −1

2[

2

3𝑡

3

2 log 𝑡 −4

9𝑡

3

2] + 𝐶

= −1

3𝑡

3

2 log 𝑡 +2

9𝑡

3

2 + 𝐶

= −1

3𝑡

3

2 [log 𝑡 −2

3] + 𝐶

= −1

3(1 +

1

𝑥2)

3

2[log (1 +

1

𝑥2) −2

3] + 𝐶 [2

Marks]

25. Evaluate the definite integral ∫ 𝑒𝑥 (1−sin 𝑥

1−cos 𝑥) 𝑑𝑥

𝜋𝜋

2

[4

Marks]

Solution:

𝐼 = ∫ 𝑒𝑥 (1−sin 𝑥

1−cos 𝑥) 𝑑𝑥

𝜋𝜋

2

= ∫ 𝑒𝑥 (1−2 sin

𝑥

2cos

𝑥

2

2 sin2𝑥

2

) 𝑑𝑥𝜋

𝜋

2

= ∫ 𝑒𝑥 (cosec2𝑥

2

2− cot

𝑥

2) 𝑑𝑥

𝜋𝜋

2

[1

Mark]

Let 𝑓(𝑥) = − cot𝑥

2

⇒ 𝑓′(𝑥) = − (−1

2cosec2 𝑥

2) =

1

2cosec2 𝑥

2 [1

Mark]

∴ 𝐼 = ∫ 𝑒𝑥(𝑓(𝑥) + 𝑓′(𝑥))𝑑𝑥𝜋

𝜋

2

[1

Mark]

= [𝑒𝑥 ∙ 𝑓(𝑥)]𝜋2

𝜋

= − [𝑒𝑥 ∙ cot𝑥

2]

𝜋2

𝜋



= − [𝑒𝜋 × cot𝜋

2− 𝑒

𝜋2 × cot

𝜋

4]

= − [𝑒𝜋 × 0 − 𝑒𝜋2 × 1]

= 𝑒𝜋

2 [1 Mark]

26. Evaluate the definite integral ∫sin 𝑥 cos 𝑥

cos4 𝑥+ sin4 𝑥𝑑𝑥

𝜋

40

[4 Marks]

Solution:

Let 𝐼 = ∫sin 𝑥 cos 𝑥

cos4 𝑥+ sin4 𝑥𝑑𝑥

𝜋

40

⇒ 𝐼 = ∫

(sin 𝑥 cos 𝑥)

cos4 𝑥

(cos4 𝑥+sin4 𝑥)

cos4 𝑥

𝑑𝑥𝜋

40

⇒ 𝐼 = ∫tan 𝑥 sec2 𝑥

1+tan4 𝑥𝑑𝑥

𝜋

40

[2

Marks]

Let tan2 𝑥 = 𝑡 ⇒ 2 tan 𝑥 sec2 𝑥 𝑑𝑥 = 𝑑𝑡


4, 𝑡 = 1

∴ 𝐼 =1

2∫

𝑑𝑡

1+𝑡2

1

0 [1

Mark]

=1

2[tan−1 𝑡]0

1

=1

2[tan−1 1 − tan−1 0]

=1

2[

𝜋

4]

=𝜋

8 [1

Mark]

27. Evaluate the definite integral ∫cos2 𝑥 𝑑𝑥

cos2 𝑥+4 sin2 𝑥

𝜋

20

[4 Marks]



Solution:

Let 𝐼 = ∫cos2 𝑥

cos2 𝑥+4 sin2 𝑥𝑑𝑥

𝜋

20

⇒ 𝐼 = ∫cos2 𝑥

cos2 𝑥+4(1−cos2 𝑥)𝑑𝑥

𝜋

20

⇒ 𝐼 = ∫cos2 𝑥

cos2 𝑥+4−4 cos2 𝑥𝑑𝑥

𝜋

20

⇒ 𝐼 =−1

3∫

4−3 cos2 𝑥−4

4−3 cos2 𝑥

𝜋

20

𝑑𝑥

⇒ 𝐼 =−1

3∫

4−3 cos2 𝑥

4−3 cos2 𝑥𝑑𝑥 +

1

3∫

4

4−3 cos2 𝑥𝑑𝑥

𝜋

20

𝜋

20

⇒ 𝐼 =−1

3∫ 1 𝑑𝑥 +

1

3∫

4 sec2 𝑥

4 sec2 𝑥−3𝑑𝑥

𝜋

20

𝜋

20

⇒ 𝐼 =−1

3[𝑥]

0

𝜋

2 +1

3∫

4 sec2 𝑥

4(1+tan2 𝑥)−3𝑑𝑥

𝜋

20

⇒ 𝐼 = −𝜋

6+

2

3 ∫

2 sec2 𝑥

1+4 tan2 𝑥

𝜋

20

𝑑𝑥 …(i) [2

Marks]

Now consider, ∫2 sec2 𝑥

1+4 tan2 𝑥𝑑𝑥

𝜋

20

Let 2 tan 𝑥 = 𝑡 ⇒ 2 sec2 𝑥 𝑑𝑥 = 𝑑𝑡


2, 𝑡 → ∞

⇒ ∫2 sec2 𝑥

1+4 tan2 𝑥𝑑𝑥 = ∫

𝑑𝑡

1+𝑡2

∞

0

𝜋

20

[1

Mark]

= [tan−1 𝑡]0∞

= [tan−1(∞) − tan−1(0)]

=𝜋

2

Hence, from (i), we get

𝐼 = −𝜋

6+

2

3[

𝜋

2] =

𝜋

3−

𝜋

6=

𝜋

6 [1

Mark]

28. Evaluate the definite integral ∫sin 𝑥+cos 𝑥

√sin 2𝑥

𝜋

3𝜋

6

𝑑𝑥 [4 Marks]



Solution:

Let 𝐼 = ∫sin 𝑥+cos 𝑥

√sin 2𝑥

𝜋

3𝜋

6

𝑑𝑥

⇒ 𝐼 = ∫(sin 𝑥+cos 𝑥)

√−(− sin 2𝑥)𝑑𝑥

𝜋

3𝜋

6

⇒ 𝐼 = ∫sin 𝑥+cos 𝑥

√−(−1+1−2 sin 𝑥 cos 𝑥)

𝜋

3𝜋

6

𝑑𝑥

⇒ 𝐼 = ∫(sin 𝑥+cos 𝑥)

√1−(sin2 𝑥+cos2 𝑥−2 sin 𝑥 cos 𝑥)𝑑𝑥

𝜋

3𝜋

6

⇒ 𝐼 = ∫(sin 𝑥+cos 𝑥)𝑑𝑥

√1−(sin 𝑥−cos 𝑥)2

𝜋

3𝜋

6

[1

Mark]

Let (sin 𝑥 − cos 𝑥) = 𝑡 ⇒ (sin 𝑥 + cos 𝑥)𝑑𝑥 = 𝑑𝑡

When 𝑥 =𝜋

6, 𝑡 = (

1−√3

2), when 𝑥 =

𝜋

3, 𝑡 = (

√3−1

2)

𝐼 = ∫𝑑𝑡

√1−𝑡2

√3−1

21−√3

2

⇒ 𝐼 = ∫𝑑𝑡

√1−𝑡2

√3−1

2

−(√3−1

2)

[1

Mark]

As 1

√1−(−𝑡)2=

1

√1−𝑡2, hence,

1

√1−𝑡2 is an even function.

As we know that if 𝑓(𝑥) is an even function, then

∫ 𝑓(𝑥)𝑑𝑥 = 2 ∫ 𝑓(𝑥)𝑑𝑥𝑎

0

𝑎

−𝑎

⇒ 𝐼 = 2 ∫𝑑𝑡

√1−𝑡2

√3−1

20

= [2 sin−1 𝑡]0

√3−1

2

= 2 sin−1 (√3−1

2) [2

Marks]


√1+𝑥−√𝑥

1

0 [4

Marks]



Solution:


√1+𝑥−√𝑥

1

0

𝐼 = ∫1

(√1+𝑥−√𝑥)×

(√1+𝑥+√𝑥)

(√1+𝑥+√𝑥)𝑑𝑥

1

0 [1

Mark]

= ∫√1+𝑥+√𝑥

1+𝑥−𝑥

1

0𝑑𝑥

= ∫ √1 + 𝑥 𝑑𝑥 + ∫ √𝑥 𝑑𝑥1

0

1

0 [1

Mark]

= [2

3(1 + 𝑥)

3

2]0

1

+ [2

3(𝑥)

3

2]0

1

[1

Mark]

=2

3[(2)

3

2 − 1] +2

3[1]

=2

3(2)

3

2

=2.2√2

3

=4√2

3 [1

Mark]

30. Evaluate the definite integral ∫sin 𝑥+cos 𝑥

9+16 sin 2𝑥𝑑𝑥

𝜋

40

[4 Marks]

Solution:

Let 𝐼 = ∫sin 𝑥+cos 𝑥

9+16 sin 2𝑥𝑑𝑥

𝜋

40

Again, let sin 𝑥 − cos 𝑥 = 𝑡 ⇒ (cos 𝑥 + sin 𝑥)𝑑𝑥 = 𝑑𝑡

When 𝑥 = 0, 𝑡 = −1, when 𝑥 =𝜋

4, 𝑡 = 0

⇒ (sin 𝑥 − cos 𝑥)2 = 𝑡2 [1 Mark]

⇒ sin2 𝑥 + cos2 𝑥 − 2 sin 𝑥 cos 𝑥 = 𝑡2

⇒ 1 − sin 2𝑥 = 𝑡2

⇒ sin 2𝑥 = 1 − 𝑡2



∴ 𝐼 = ∫𝑑𝑡

9+16(1−𝑡2)

0

−1 [1

Mark]

= ∫𝑑𝑡

9+16−16𝑡2

0

−1

= ∫𝑑𝑡

25−16𝑡2 = ∫𝑑𝑡

(5)2−(4𝑡)2

0

−1

0

−1

=1

4[

1

2(5)log |

5+4𝑡

5−4𝑡|]

−1

0

=1

40[log(1) − log |

1

9|]

=1

40log 9 [2

Marks]

31. Evaluate the definite integral ∫ sin 2𝑥 tan−1(sin 𝑥) 𝑑𝑥𝜋

20

[6 Marks]

Solution:

Let 𝐼 = ∫ sin 2𝑥 tan−1(sin 𝑥) 𝑑𝑥𝜋

20

= ∫ 2 sin 𝑥 cos 𝑥 tan−1(sin 𝑥) 𝑑𝑥𝜋

20

Again, let sin 𝑥 = 𝑡 ⇒ cos 𝑥 𝑑𝑥 = 𝑑𝑡


2, 𝑡 = 1

⇒ 𝐼 = 2 ∫ 𝑡 tan−1(𝑡) 𝑑𝑡1

0 …(i) [2

Marks]

Now, take ∫ 𝑡 ∙ tan−1 𝑡 𝑑𝑡 = tan−1 𝑡 ∙ ∫ 𝑡 𝑑𝑡 − ∫ {𝑑

𝑑𝑡(tan−1 𝑡) ∫ 𝑡 𝑑𝑡} 𝑑𝑡

= tan−1 𝑡 ∙𝑡2

2− ∫

1

1+𝑡2 ∙𝑡2

2𝑑𝑡

=𝑡2 tan−1 𝑡

2−

1

2∫

𝑡2+1−1

1+𝑡2 𝑑𝑡

=𝑡2 tan−1 𝑡

2−

1

2∫ 1 𝑑𝑡 +

1

2∫

1

1+𝑡2 𝑑𝑡

=𝑡2 tan−1 𝑡

2−

1

2∙ 𝑡 +

1

2tan−1 𝑡

⇒ ∫ 𝑡 ∙ tan−1 𝑡 𝑑𝑡 = [𝑡2∙tan−1 𝑡

2−

𝑡

2+

1

2tan−1 𝑡]

0

11

0 [2

Marks]

=1

2[𝜋

4− 1 +

𝜋

4]



=1

2[𝜋

2− 1] =

𝜋

4−

1

2


𝐼 = 2 [𝜋

4−

1

2] =

𝜋

2− 1 [2

Marks]

32. Evaluate the definite integral ∫𝑥 tan 𝑥

sec 𝑥+tan 𝑥𝑑𝑥

𝜋

0 [6 Marks]

Solution:

Let 𝐼 = ∫𝑥 tan 𝑥


𝜋

0 …(i)

𝐼 = ∫ {(𝜋−𝑥) tan(𝜋−𝑥)

sec(𝜋−𝑥)+tan(𝜋−𝑥)} 𝑑𝑥

𝜋

0(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)

𝑎

0𝑑𝑥

𝑎

0)

⇒ 𝐼 = ∫ {−(𝜋−𝑥) tan 𝑥

−(sec 𝑥+tan 𝑥)} 𝑑𝑥

𝜋

0

⇒ 𝐼 = ∫(𝜋−𝑥) tan 𝑥


𝜋

0 …(ii) [2

Marks]


2𝐼 = ∫𝜋 tan 𝑥


𝜋

0 [1

Mark]

⇒ 2𝐼 = 𝜋 ∫sin 𝑥

cos 𝑥1

cos 𝑥+

sin 𝑥

cos 𝑥

𝑑𝑥𝜋

0

⇒ 2𝐼 = 𝜋 ∫sin 𝑥+1−1

1+sin 𝑥𝑑𝑥

𝜋

0

⇒ 2𝐼 = 𝜋 ∫ 1 ∙ 𝑑𝑥 − 𝜋 ∫1

1+sin 𝑥

𝜋

0𝑑𝑥

𝜋

0

⇒ 2𝐼 = 𝜋[𝑥]0𝜋 − 𝜋 ∫

1−sin 𝑥

cos2 𝑥𝑑𝑥

𝜋

0

⇒ 2𝐼 = 𝜋2 − 𝜋 ∫ (sec2 𝑥 − tan 𝑥 sec 𝑥)𝑑𝑥𝜋

0 [1

Mark]

⇒ 2𝐼 = 𝜋2 − 𝜋[tan 𝑥 − sec 𝑥]0𝜋

⇒ 2𝐼 = 𝜋2 − 𝜋[tan 𝜋 − sec 𝜋 − tan 0 + sec 0]

⇒ 2𝐼 = 𝜋2 − 𝜋[0 − (−1) − 0 + 1]

⇒ 2𝐼 = 𝜋2 − 2𝜋



⇒ 2𝐼 = 𝜋(𝜋 − 2)

⇒ 𝐼 =𝜋

2(𝜋 − 2) [2

Marks]

33. Evaluate the definite integral ∫ [|𝑥 − 1| + |𝑥 − 2| + |𝑥 − 3|]𝑑𝑥4

1 [6

Marks]

Solution:

Let 𝐼 = ∫ [|𝑥 − 1| + |𝑥 − 2| + |𝑥 − 3|]𝑑𝑥4

1

⇒ 𝐼 = ∫ |𝑥 − 1|𝑑𝑥 + ∫ |𝑥 − 2|𝑑𝑥 + ∫ |𝑥 − 3|𝑑𝑥4

1

4

1

4

1

𝐼 = 𝐼1 + 𝐼2 + 𝐼3 …(i) [1 Mark]

where, 𝐼1 = ∫ |𝑥 − 1|𝑑𝑥, 𝐼2 = ∫ |𝑥 − 2|𝑑𝑥4

1

4

1 and 𝐼3 = ∫ |𝑥 − 3|𝑑𝑥

4

1 [1

Mark]

𝐼1 = ∫ |𝑥 − 1|𝑑𝑥4

1

(𝑥 − 1) ≥ 0 for 1 ≤ 𝑥 ≤ 4

∴ 𝐼1 = ∫ (𝑥 − 1)𝑑𝑥4

1

⇒ 𝐼1 = [𝑥2

2− 𝑥]

1

4

⇒ 𝐼1 = [8 − 4 −1

2+ 1] =

9

2 …(ii) [1

Mark]

𝐼2 = ∫ |𝑥 − 2|𝑑𝑥4

1

𝑥 − 2 ≥ 0 for 2 ≤ 𝑥 ≤ 4 and 𝑥 − 2 ≤ 0 for 1 ≤ 𝑥 ≤ 2

∴ 𝐼2 = ∫ (2 − 𝑥)𝑑𝑥 + ∫ (𝑥 − 2)𝑑𝑥4

2

2

1

⇒ 𝐼2 = [2𝑥 −𝑥2

2]

1

2

+ [𝑥2

2− 2𝑥]

2

4

⇒ 𝐼2 = [4 − 2 − 2 +1

2] + [8 − 8 − 2 + 4]

⇒ 𝐼2 =1

2+ 2 =

5

2 …(iii) [1

Mark]



𝐼3 = ∫ |𝑥 − 3|𝑑𝑥4

1

𝑥 − 3 ≥ 0 for 3 ≤ 𝑥 ≤ 4 and 𝑥 − 3 ≤ 0 for 1 ≤ 𝑥 ≤ 3

∴ 𝐼3 = ∫ (3 − 𝑥)𝑑𝑥 + ∫ (𝑥 − 3)𝑑𝑥4

3

3

1

⇒ 𝐼3 = [3𝑥 −𝑥2

2]

1

3

+ [𝑥2

2− 3𝑥]

3

4

⇒ 𝐼3 = [9 −9

2− 3 +

1

2] + [8 − 12 −

9

2+ 9]

⇒ 𝐼3 = [6 − 4] + [1

2] =

5

2 …(iv) [1

Mark]

From equations (i), (ii), (iii) and (iv), we get

𝐼 =9

2+

5

2+

5

2=

19

2 [1

Mark]

34. Prove that ∫𝑑𝑥

𝑥2(𝑥+1)=

2

3+ log

2

3

3

1 [6

Marks]

Solution:


𝑥2(𝑥+1)

3

1

Again, let 1

𝑥2(𝑥+1)=

𝐴

𝑥+

𝐵

𝑥2 +𝐶

𝑥+1 [1

Mark]

⇒ 𝐼 = 𝐴𝑥(𝑥 + 1) + 𝐵(𝑥 + 1) + 𝐶(𝑥2)

⇒ 𝐼 = 𝐴𝑥2 + 𝐴𝑥 + 𝐵𝑥 + 𝐵 + 𝐶𝑥2 [1 Mark]

After equating the coefficients of 𝑥2, 𝑥 and constant term, we get

𝐴 + 𝐶 = 0

𝐴 + 𝐵 = 0

𝐵 = 1

after solving these equations, we get

𝐴 = −1, 𝐶 = 1 and 𝐵 = 1 [1 Mark]



∴1

𝑥2(𝑥 + 1)=

−1

𝑥+

1

𝑥2+

1

(𝑥 + 1)

⇒ 𝐼 = ∫ {−1

𝑥+

1

𝑥2 +1

(𝑥+1)} 𝑑𝑥

3

1 [1

Mark]

= [− log 𝑥 −1

𝑥+ log(𝑥 + 1)]

1

3

= [log (𝑥 + 1

𝑥) −

1

𝑥]

1

3

= log (4

3) −

1

3− log (

2

1) + 1

= log 4 − log 3 − log 2 +2

3

= log 2 − log 3 +2

3

= log (2

3) +

2

3

Thus, its proved [2 Marks]

35. Prove that ∫ 𝑥𝑒𝑥𝑑𝑥 = 11

0 [2

Marks]

Solution:

Consider 𝐼 = ∫ 𝑥𝑒𝑥1

0𝑑𝑥


𝐼 = 𝑥 ∫ 𝑒𝑥𝑑𝑥 − ∫ {(𝑑

𝑑𝑥(𝑥)) ∫ 𝑒𝑥𝑑𝑥} 𝑑𝑥

1

0

1

0 [1

Mark]

= [𝑥𝑒𝑥]01 − ∫ 𝑒𝑥𝑑𝑥

1

0

= [𝑥𝑒𝑥]01 − [𝑒𝑥]0

1

= 𝑒 − 𝑒 + 1

= 1



Therefore, it’s proved [1 Mark]

36. Prove that ∫ 𝑥17 cos4 𝑥𝑑𝑥 = 01

−1 [2

Marks]

Solution:

Let 𝐼 = ∫ 𝑥171

−1cos4 𝑥𝑑𝑥

Again, let 𝑓(𝑥) = 𝑥17 cos4 𝑥

⇒ 𝑓(−𝑥) = (−𝑥)17 cos4(−𝑥) = −𝑥17 cos4 𝑥 = −𝑓(𝑥)

Hence, 𝑓(𝑥) is an odd function. [1 Mark]

We know that if 𝑓(𝑥) is an odd function, then ∫ 𝑓(𝑥)𝑑𝑥 = 0𝑎

−𝑎

∴ 𝐼 = ∫ 𝑥17 cos4 𝑥𝑑𝑥 = 01

−1

Hence, it’s proved. [1 Mark]

37. Prove that ∫ sin3𝑥 𝑑𝑥 =2

3

𝜋

20

[4

Marks]

Solution:

Consider 𝐼 = ∫ sin3 𝑥 𝑑𝑥𝜋

20

𝐼 = ∫ sin2 𝑥 ∙ sin 𝑥 𝑑𝑥𝜋

20

= ∫ (1 − cos2 𝑥)𝜋

20

sin 𝑥 𝑑𝑥

= ∫ sin 𝑥 𝑑𝑥 − ∫ cos2𝑥 ∙ sin 𝑥 𝑑𝑥𝜋

20

𝜋

20

[2

Marks]

= [− cos 𝑥]0

𝜋2 + [

cos3 𝑥

3]

0

𝜋2



= 1 +1

3[−1] = 1 −

1

3=

2

3

Thus, its proved [2 Marks]

38. Prove that ∫ 2 tan3 𝑥𝑑𝑥 = 1 − log 2𝜋

40

[4 Marks]

Solution:

Consider 𝐼 = ∫ 2 tan3 𝑥𝑑𝑥𝜋

40

𝐼 = 2 ∫tan2 𝑥 tan 𝑥 𝑑𝑥

𝜋

40

= 2 ∫(sec2 𝑥 − 1) tan 𝑥 𝑑𝑥

𝜋4

0

= 2 ∫ sec2 𝑥 tan 𝑥 𝑑𝑥 − 2 ∫ tan 𝑥 𝑑𝑥𝜋

40

𝜋

40

[2

Marks]

= 2 [tan2 𝑥

2]

0

𝜋4

+ 2[log cos 𝑥]0

𝜋4

= 1 + 2 [log cos𝜋

4− log cos 0]

= 1 + 2 [log1

√2− log 1]

= 1 − log 2 − log 1 = 1 − log 2

Hence, it’s proved. [2 Marks]

39. Prove that ∫ sin−1 𝑥𝑑𝑥 =𝜋

2− 1

1

0 [4

Marks]

Solution:



Let’s consider 𝐼 = ∫ sin−1 𝑥𝑑𝑥1

0

⇒ 𝐼 = ∫ sin−1 𝑥 ∙ 1 ∙ 𝑑𝑥1

0 [1

Mark]


𝐼 = [sin−1 𝑥 ∙ 𝑥]01 − ∫

1

√1 − 𝑥2∙ 𝑥𝑑𝑥

1

0

= [𝑥 sin−1 𝑥]01 +

1

2∫

(−2𝑥)


1

0 [1

Mark]

Again Let 1 − 𝑥2 = 𝑡 ⇒ −2𝑥 𝑑𝑥 = 𝑑𝑡

When 𝑥 = 0, 𝑡 = 1, when 𝑥 = 1, 𝑡 = 0

𝐼 = [𝑥 sin−1 𝑥]01 +

1

2∫

𝑑𝑡

√𝑡

0

1 [1

Mark]

= [𝑥 sin−1 𝑥]01 +

1

2[2√𝑡]

1

0

= sin−1(1) + [−√1]

=𝜋

2− 1

Hence, it’s proved. [1 Mark]

40. Evaluate ∫ 𝑒2−3𝑥𝑑𝑥1

0 as a limit of a sum. [6

Marks]

Solution:

Let’s consider 𝐼 = ∫ 𝑒2−3𝑥𝑑𝑥1

0

We know that,


1

𝑛[𝑓(𝑎) + 𝑓(𝑎 + ℎ) + ⋯ + 𝑓(𝑎 + (𝑛 − 1)ℎ)]

𝑏

𝑎 [1 Mark]

Where, ℎ =𝑏−𝑎

𝑛

For, 𝑎 = 0, 𝑏 = 1 and 𝑓(𝑥) = 𝑒2−3𝑥



⇒ ℎ =1−0

𝑛=

1

𝑛 [1

Mark]

∴ ∫ 𝑒2−3𝑥𝑑𝑥 = (1 − 0) lim𝑛→∞

1

𝑛[𝑓(0) + 𝑓(0 + ℎ) + ⋯ + 𝑓(0 + (𝑛 − 1)ℎ)]

4

0 [1

Mark]

= lim𝑛→∞

1

𝑛[𝑒2 + 𝑒2−3ℎ + ⋯ 𝑒2−3(𝑛−1)ℎ]

= lim𝑛→∞

1

𝑛[𝑒2{1 + 𝑒−3ℎ + 𝑒−6ℎ + 𝑒−9ℎ + ⋯ 𝑒−3(𝑛−1)ℎ}]

= lim𝑛→∞

1

𝑛[𝑒2 {

1−(𝑒−3ℎ)𝑛

1−(𝑒−3ℎ)}] [1

Mark]

= lim𝑛→∞

1

𝑛[𝑒2 {

1 − 𝑒−3𝑛

×𝑛

1 − 𝑒−3𝑛

}]

= lim𝑛→∞

1

𝑛[𝑒2(1 − 𝑒−3)

1 − 𝑒−3𝑛

]

= 𝑒2(𝑒−3 − 1) lim𝑛→∞

1

𝑛[

1

𝑒−3𝑛 − 1

]

= 𝑒2(𝑒−3 − 1) lim𝑛→∞

(−1

3) [

−3𝑛

𝑒−3𝑛 − 1

]

=−𝑒2(𝑒−3 − 1)

3lim

𝑛→∞[

−3𝑛

𝑒−3𝑛 − 1

]

=−𝑒2(𝑒−3 − 1)

3(1) [ lim

𝑛→∞

𝑥

𝑒𝑥 − 1]

=−𝑒−1 + 𝑒2

3

=1

3(𝑒2 −

1

𝑒) [2

Marks]

41. ∫𝑑𝑥

𝑒𝑥+𝑒−𝑥 is equal to [2

Marks]

(A) tan−1(𝑒𝑥) + 𝐶



(B) tan−1(𝑒−𝑥) + 𝐶

(C) log(𝑒𝑥 − 𝑒−𝑥) + 𝐶

(D) log(𝑒𝑥 + 𝑒−𝑥) + 𝐶

Solution:

(A)

Consider 𝐼 = ∫𝑑𝑥

𝑒𝑥+𝑒−𝑥 𝑑𝑥 = ∫𝑒𝑥

𝑒2𝑥+1𝑑𝑥

Again, let 𝑒𝑥 = 𝑡 ⇒ 𝑒𝑥𝑑𝑥 = 𝑑𝑡

∴ 𝐼 = ∫𝑑𝑡

1+𝑡2 [1

Mark]

= tan−1 𝑡 + 𝐶

= tan−1(𝑒𝑥) + 𝐶

Therefore, (𝐴) is the correct answer. [1 Mark]

42. ∫cos 2𝑥

(sin 𝑥+cos 𝑥)2 𝑑𝑥 is equal to [4

Marks]

(A) −1

sin 𝑥+cos 𝑥+ 𝐶

(B) log|sin 𝑥 + cos 𝑥| + 𝐶

(C) log|sin 𝑥 − cos 𝑥| + 𝐶

(D) 1

(sin 𝑥+cos 𝑥)2

Solution:

(B)

Consider 𝐼 =cos 2𝑥

(cos 𝑥+sin 𝑥)2

𝐼 = ∫cos2 𝑥 − sin2 𝑥

(cos 𝑥 + sin 𝑥)2𝑑𝑥

= ∫(cos 𝑥 + sin 𝑥)(cos 𝑥 − sin 𝑥)

(cos 𝑥 + sin 𝑥)2𝑑𝑥



= ∫cos 𝑥−sin 𝑥

cos 𝑥 + sin 𝑥𝑑𝑥 [2

Marks]

Also, let cos 𝑥 + sin 𝑥 = 𝑡 ⇒ (cos 𝑥 − sin 𝑥)𝑑𝑥 = 𝑑𝑡 [1 Mark]

∴ 𝐼 = ∫𝑑𝑡

𝑡

= log|𝑡| + 𝐶

= log|cos 𝑥 + sin 𝑥| + 𝐶

Hence, (𝐵) is the correct answer. [1 Mark]

43. If 𝑓(𝑎 + 𝑏 − 𝑥) = 𝑓(𝑥), then ∫ 𝑥 𝑓(𝑥)𝑏

𝑎𝑑𝑥 is equal to [4 Marks]

(A) 𝑎+𝑏

2∫ 𝑓(𝑏 − 𝑥)

𝑏

𝑎𝑑𝑥

(B) 𝑎+𝑏

2∫ 𝑓(𝑏 + 𝑥)𝑑𝑥

𝑏

𝑎

(C) 𝑏−𝑎

2∫ 𝑓(𝑥)𝑑𝑥

𝑏

𝑎

(D) 𝑎+𝑏

2 ∫ 𝑓(𝑥)

𝑏

𝑎𝑑𝑥

Solution:

(D)

Consider 𝐼 = ∫ 𝑥 𝑓(𝑥)𝑏

𝑎𝑑𝑥 …(i)

𝐼 = ∫ (𝑎 + 𝑏 − 𝑥)𝑓(𝑎 + 𝑏 − 𝑥)𝑑𝑥𝑏

𝑎(∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 + 𝑏 − 𝑥)𝑑𝑥

𝑏

𝑎

𝑏

𝑎) [1

Mark]

⇒ 𝐼 = ∫(𝑎 + 𝑏 − 𝑥)𝑓(𝑥)𝑑𝑥

𝑏

𝑎

⇒ 𝐼 = (𝑎 + 𝑏) ∫ 𝑓(𝑥)𝑑𝑥 − 𝐼𝑏

𝑎 [Using (i)] [1

Mark]

⇒ 𝐼 + 𝐼 = (𝑎 + 𝑏) ∫ 𝑓(𝑥)𝑑𝑥

𝑏

𝑎



⇒ 2𝐼 = (𝑎 + 𝑏) ∫ 𝑓(𝑥)𝑑𝑥

𝑏

𝑎

⇒ 𝐼 = (𝑎 + 𝑏

2) ∫ 𝑓(𝑥)𝑑𝑥

𝑏

𝑎

Hence, (𝐷) is the correct answer. [2 Marks]

44. The value of ∫ tan−1 (2𝑥−1

1+𝑥−𝑥2) 𝑑𝑥1

0 is

[4 Marks]

(A) 1

(B) 0

(C) −1

(D) 𝜋

4

Solution:

(B)

Consider 𝐼 = ∫ tan−1 (2𝑥−1

1+𝑥−𝑥2) 𝑑𝑥1

0

⇒ 𝐼 = ∫ tan−1 (𝑥−(1−𝑥)

1+𝑥(1−𝑥)) 𝑑𝑥

1

0

⇒ 𝐼 = ∫ [tan−1 𝑥 − tan−1(1 − 𝑥)]𝑑𝑥1

0……(i) [1

Mark]

⇒ 𝐼 = ∫ [tan−1(1 − 𝑥) − tan−1(1 − 1 + 𝑥)]𝑑𝑥1

0

⇒ 𝐼 = ∫ [tan−1(1 − 𝑥) − tan−1(𝑥)]𝑑𝑥1

0

⇒ 𝐼 = ∫ [tan−1(1 − 𝑥) − tan−1(𝑥)]𝑑𝑥1

0 …(ii) [1

Mark]

Adding (i) and (ii), we get,

2𝐼 = ∫ (tan−1 𝑥 − tan−1(1 − 𝑥) + tan−1(1 − 𝑥) − tan−1 𝑥)𝑑𝑥1

0

⇒ 2𝐼 = 0

⇒ 𝐼 = 0



Hence, (𝐵) is the correct answer. [2 Marks]

CBSE NCERT Solutions for Class 12 Maths Chapter 07 · CBSE NCERT Solutions for Class 12 Maths...

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Transcript of CBSE NCERT Solutions for Class 12 Maths Chapter 07 · CBSE NCERT Solutions for Class 12 Maths...