2 Mahesh Tutorials Science · 1 x + 2 ∫ dx = 2 2 2 2 0 0 1 dx 8 dx x 2 ... 1 1 2 1 0 0 1 1 x logx...
36
G ROUP (A)- HOME WORK PROBLEMS 1) 1 2 –1 (x + 1) (2x + 3) dx ∫ Ans. = 1 2 –1 (x +1) 1(2x + 3) dx + ∫ = 1 3 2 –1 (2x + 3x + 2x + 3) dx ∫ =2 1 3 –1 x dx + 3 ∫ 1 2 –1 x dx + 2 ∫ 1 –1 x dx + 3 ∫ 1 –1 1 dx ∫ = 1 4 –1 4 x 2 + 1 3 –1 3 x 3 + 1 2 –1 2 x 2 + [ ] 1 –1 3x dx = 1 2 (1 – 1) + 1(1 + 1)+ [1– 1]+ 3 [ 1 + 1] = 2 + 3(2) = 2 + 6 = 8 2) 2 2 2 0 x – 4 x + 4 ∫ dx = ( ) 2 2 2 0 x + 4 8 dx x + 4 - ∫ Ans. = 2 2 2 0 x + 4 dx x + 4 ∫ – 8 2 2 2 0 1 x + 2 ∫ dx = 2 2 2 2 0 0 1 dx 8 dx x 2 - + ∫ ∫ = [ ] 2 0 x – 2 –1 0 8 x tan 2 2 = (2 – 0) – 4 –1 2 tan 0 2 - = 2 – 4 – 0 4 π = 2 – π 3) 1 4 1/2 2 –3/2 1 13 x + 3x + 4 ∫ dx Ans. 1/2 2 –3/2 1 dx 3 9 13 9 4 x + 2x . 2 4 4 4 = + + - ∫ 1/2 2 2 –3/2 1 1 dx 4 3 x + + 1 2 = ∫ = 1 4 1/2 –1 –3/2 3 tan x + 2 = 1 4 [tan –1 (2) – 0] = 1 4 tan –1 (2) 4) ( )( ) 2 1 dx x 1 x 2 + + ∫ Ans. Let, I = ( )( ) 2 1 dx x 1 x 2 + + ∫ 2 1 1 1 x 1 x 2 = - + + ∫ dx ( ) 2 1 log x 1 log x 2 = + - + 2 1 x 1 log x 2 + = + 3 2 log log 4 3 = - 3 2 log 4 3 = ÷ 9 log 8 = Definite Integration
Transcript of 2 Mahesh Tutorials Science · 1 x + 2 ∫ dx = 2 2 2 2 0 0 1 dx 8 dx x 2 ... 1 1 2 1 0 0 1 1 x logx...
GROUP (A)-HOME WORK PROBLEMS
1)
12
–1
(x + 1) (2x + 3) dx∫
Ans. =
12
–1
(x +1) 1(2x + 3) dx + ∫
=
13 2
–1
(2x + 3x + 2x + 3) dx∫
=2
13
–1
x dx + 3∫1
2
–1
x dx + 2∫1
–1
x dx + 3∫1
–1
1 dx∫
=1
4
–1
4x
2 +
13
–1
3x
3 +
12
–1
2x
2 + [ ]
1–1
3 x dx
= 1
2(1 – 1) + 1(1 + 1)+ [1– 1]+ 3 [ 1 + 1]
= 2 + 3(2) = 2 + 6 = 8
2)
2 2
20
x – 4
x + 4∫ dx =
( )22
20
x + 4 8dx
x + 4
−
∫
Ans. =
2 2
20
x + 4dx
x + 4∫ – 8
2
2 20
1
x + 2∫ dx
=
2 2
2 20 0
1dx 8 dx
x 2−
+∫ ∫
= [ ]20
x –
2–1
0
8 xtan
2 2
= (2 – 0) – 4–1 2
tan 02
−
= 2 – 4 – 04
π
= 2 – π
3)1
4
1/2
2–3/2
1
13x + 3x +
4
∫ dx
Ans.
1/2
2–3/2
1 dx
3 9 13 94x + 2x .
2 4 4 4
=
+ + −
∫
1/2
22–3/2
1 1 dx
4 3x + + 1
2
=
∫
= 1
4
1/2–1
–3/2
3tan x +
2
= 1
4 [tan–1 (2) – 0]
= 1
4 tan–1 (2)
4) ( ) ( )
2
1
dx
x 1 x 2+ +∫
Ans. Let, I = ( ) ( )
2
1
dx
x 1 x 2+ +∫
2
1
1 1
x 1 x 2
= − + + ∫ dx
( )2
1log x 1 log x 2= + − +
2
1
x 1log
x 2
+=
+
3 2
log log4 3
= −
3 2
log4 3
= ÷
9
log8
=
Definite Integration
Definite Integration
2 Mahesh Tutorials Science
5)
2
20
5x + 1
x + 4∫ dx
Ans. = 5
2
20
x
x + 4∫ dx +
2
20
1
x + 4∫ dx
2 2
2 20 0
5 2x 1dx dx
2 x + 4 x + 4= +∫ ∫
= 5
2
22
0log |x + 4| +
1
2
2–1
0
xtan
2
=5
2[log(22 + 4)– log4]+
1
2
–1 2tan 0
2
−
= 5
2
8log
4
+ 1
2 – 04
π
= 8
π +
5
2 log 2
6)
1
20
1
– (x – 2x – 3)∫ dx
Ans. =
1
20
dx
– (x – 2x +1– 4)∫
=
1
2 20
1
2 – (x – 1)∫ dx
11
0
x 1sin
2
− − =
= 0 6
π +
= 6
π+
7)
5
3
1
x + 4 + x – 2∫ dx
Ans. = ( ) ( )
5
3
x 4 x 2dx
x 4 x 2 x 4 x 2
+ − −
+ + − + − −∫
= 1
6
5 5
3 3
x + 4 – x – 2 dx ∫ ∫
= 2
18
5 53/2 3/2
3 3(x + 4) – (x – 2)
= 1
9 { }27 – 7 7 – 3 3 – 1
= 1
9 27 – 7 7 – 3 3 + 1
=1
928 – 7 7 – 3 3
8)
/2
0
dx
1 + cos x
π
∫
Ans. = 1
2
/22
0
xsec
2
π
∫
= 1
2
/2
0
xtan
21
2
π
1
2 tan 02 4
π = × −
= 1
9)
/4
/4
dx
1 – sin x
π
−π
∫
Ans. =
/4
2– /4
1 + sin x
cos x
π
π
∫
( )/4
2
/4
sec x sec x tan x dx
π
π−
= +∫
= [ ]/4– /4
tan x + sec xπ
π
= tan sec tan sec4 4 4 4
π π π π + − − +
= ( )1 2 1 2+ − − +
=1 2 1 2+ + −
= 2
10)
/2 2
2/3
sin x
(1 – cos x)
π
π
∫ dx
Ans. =
2/2
2/3
x x2 sin cos 2 2
x2 sin 2
π
π
∫ dx
=
/22
/3
xcot
2
π
π
∫ dx
Definite Integration
Mahesh Tutorials Science 3
=
/22
/3
xcosec – 1
2
π
π
∫ dx
[ ]/2
/3
/3
/2x
cot2 x x
12
π
π
π
π
= − −
= – 2 1 – 3 –
– 2 3
π π
= –2 (1 – 3) – 6
π
= –2 + 2 3 – 6
π
BOARD PROBLEMS
1) ( )2
2
0
3x 2x 1 dx+ −∫
Ans. I = ( )2
2
0
3x 2x 1 dx+ −∫
=
2 2 22
0 0 0
3 x dx 2 x dx dx+ −∫ ∫ ∫
= ( ) ( ) ( )2
0
2 23 2
0 0
23 x x x
2+ −
= 8 + 4 – 2 = 10
2)
1 2
0
x 3x 2dx
x
+ +∫
Ans. I =
1 2
0
x 3x 2dx
x
+ +∫
=
1 1 12
0 0 0
x x dxdx 3 dx 2
x x x+ +∫ ∫ ∫
=
1 1 13 12 2
0 0 0
dxx dx 3 x dx 2
x+ +∫ ∫ ∫
= ( )1 1
5 3 12 2
00 0
2 2x 3 x 2.2 x
5 3
+ × +
= ( ) ( )21 2 1 4
5+ +
=2
65
+
=32
5
3)
1
0
xdx
x 1+∫
Ans. I =
1
0
xdx
x 1+∫
=( )1
0
x 1 1dx
x 1
+ −
+∫
=
1
0
11 dx
x 1
−
+ ∫
=
1 1
0 0
dxdx
x 1−
+∫ ∫
= ( )11
0 0x log x 1− +
= 1 – log 2
4)
1x
0
2 dx∫
Ans. I =
1x
0
2 dx∫
=
1x
0
2
log 2
= ( )1 012 2
log 2−
= ( )1
2 1log 2
−
=1
log 2
Definite Integration
4 Mahesh Tutorials Science
5)
1
20
dx
1 x+∫
Ans. I =
1
20
dx
1 x+∫
= ( )1
1
0tan x−
= tan–1 (1) – 0
=4
π
6)
1 2
20
xdx
1 x+∫
Ans. I =
1 2
20
xdx
1 x+∫
=( )21
20
x 1 1dx
1 x
+ −
+∫
=
1 1
20 0
dxdx
1 x−
+∫ ∫
= ( ) ( )11
0 0
1x tan x−−
= ( ){ }11 tan 1−−
=14
π−
7)
1 2
20
x 1dx
x 1
−
+∫
Ans. I =
1 2
20
x 1dx
x 1
−
+∫
=( )21
20
x 1 2dx
x 1
+ −
+∫
=
1 1
20 0
dxdx 2
x 1−
+∫ ∫
= ( ) ( )1
0
11
0x 2 tan x−
−
= ( ){ }11 2 tan 1−−
= 1 2.4
π−
=12
π−
8)
1 2
20
1 xdx
1 x
−
+∫
Ans. I =
1 2
20
1 xdx
1 x
−
+∫
= ( )21
20
2 x 1dx
1 x
− +
+∫
=
1 1
20 0
dx2 dx
1 x−
+∫ ∫
= ( ) ( )1
0
11
02 tan x x−
=1 2.4
π−
= 12
π−
9)
1
0
dx
x 3 x 1+ − +∫
Ans. I=
1
0
dx
x 3 x 1+ − +∫
=
( )
( ) ( )
1
0
x 3 x 1dx
x 3 x 1 x 3 x 1
+ + +
+ − + + + +∫
= ( ) ( )
1
0
x 3 x 1dx
x 3 x 1
+ + +
+ − +∫
=
1 1
0 0
1x 3 dx x 1dx
2
+ + + ∫ ∫
= ( ) ( )1 1
3 32 2
0 0
1 2 1 2x 3 x 1
2 3 2 3
× + + × +
=
3 3 3 32 2 2 21
4 3 2 13
− + −
=1
8 3 3 2 2 13 − + −
Definite Integration
Mahesh Tutorials Science 5
=17 3 3 2 2
3 − +
10)
1 2
20
2 xdx
1 x
−
+∫
Ans. I =
1 2
20
2 xdx
1 x
−
+∫
∴∴∴∴ I =
( )( )
21
20
3 x 1dx
1 x
− +
+∫
∴∴∴∴ I =
1
20
31 dx
1 x
−
+ ∫
∴∴∴∴ I =
1 1
20 0
13 dx dx1 x
−+
∫ ∫
∴∴∴∴ I = ( ) ( )1
0
11
03 tan x x−
−
∴∴∴∴ I = 3. 14
π−
∴∴∴∴ I = 3
14
π−
11)
1
20
x 4dx
x 5
+
+∫
Ans. I =
1
20
x 4dx
x 5
+
+∫
∴∴∴∴ I =
1 1
2 20 0
x dxdx 4
x 5 x 5+
+ +∫ ∫
∴∴∴∴ I = ( ) ( )
1 1
2 220 0
1 2x dxdx 4
2 x 5 x 5
++ +
∫ ∫
∴∴∴∴ I =
112 1
0 0
1 1 xlog x 5 4. tan
2 5 5
− + +
∴∴∴∴ I = [ ]1
1
0
1 1 xlog 6 log 5 4. tan
2 5 5
− − +
∴∴∴∴ I = [ ] 11 4 1log 6 log 5 tan
2 5 5
− − +
∴∴∴∴ I = 11 6 4 1
log tan2 5 5 5
− +
12)
5
22
dx
5 4x x+ −∫
Ans. I =
5
22
dx
5 4x x+ −∫
∴∴∴∴ I =
5
22
dx
9 4 4x x− + −∫
∴∴∴∴ I = ( )
5
22
dx
9 x 4x 4− − +∫
∴∴∴∴ I =
51
2
x 2sin
3
− −
∴∴∴∴ I = ( )1 15 2sin sin 0
3
− −− −
∴∴∴∴ I = 2
π
13) ( ) ( )
3
2
x dx
x 3 x 2+ +∫
Ans. Let ( ) ( ) ( ) ( )
x A B
x 3 x 2 x 3 x 2= +
+ + + +
∴∴∴∴ x = A(x + 2) + B(x + 3)
∴∴∴∴ x = (A + B) x + (2A + 3B)
Compairing co-efficients of x and constants
from both sides, we have
A + B = 1 } × 22A + 3B = 0
2A 2B 2
2A 3B 0
B 2
− −−
+ =
+ =
− =
∴∴∴∴ B = –2
∴∴∴∴ A = 1 – B
∴∴∴∴ A = 1 –(–2)
∴∴∴∴ A = 3
∴∴∴∴ I =
3
2
3 2dx
x 3 x 2
−
+ + ∫
Definite Integration
6 Mahesh Tutorials Science
∴∴∴∴ I = 3 3
2 23 log x 3 2 log x 2 + − +
∴∴∴∴ I = [ ] [ ]3 log 6 log5 2 log5 log 4− − −
∴∴∴∴ I = 6 5
3 log 2log5 4
−
∴∴∴∴ I =
3 26 5
log log5 4
−
∴∴∴∴ I =
3 2
3 2
6 4log
5 5
×
=3456
log3125
14)
2
21
dx
x 2x 2− +∫
Ans. I =
2
21
dx
x 2x 2− +∫
∴∴∴∴ I = ( )
2
2 21
dx
x 1 1− +∫
∴∴∴∴ I =
211 x 1
tan1 1
− −
∴∴∴∴ I = tan–1 2 – tan–1 0
∴∴∴∴ I = tan–1 2
15)
3
22
dx
x 5x 6+ +∫
∴∴∴∴ I =
3
22
dx
x 5x 6+ +∫
∴∴∴∴ I = ( ) ( )
3
2
dx
x 2 x 3+ +∫
∴∴∴∴ I =
3
2
1 1dx
x 2 x 3
−
+ + ∫
∴∴∴∴ I =
3 3
2 2
1 dxdx
x 2 x 3−
+ +∫ ∫
∴∴∴∴ I = 3 3
2 2log x 2 log x 3 + − +
∴∴∴∴ I = [ ] ( )log5 log 4 log 6 log5− − −
∴∴∴∴ I = 2 log 5 – (log 4 + log 6)
∴∴∴∴ I = log 52 – log 24
∴∴∴∴ I = log25
24
16)
22
0
sin x dx
π
∫
Ans. I =
22
0
sin x dx
π
∫
∴∴∴∴ I = ( )2
0
11 cos2x dx
2
π
−∫
∴∴∴∴ I =
2 2
0 0
1dx cos 2x dx
2
π π
−
∫ ∫
∴∴∴∴ I = ( )2
0
2
0
1 sin2xx
2 2
π π −
∴∴∴∴ I = ( )1 1
sin sin02 2 2
π − π −
∴∴∴∴ I = 4
π
17)
32
6
sin x dx
π
π∫
Ans. I =
32
6
sin x dx3 6
π
π
π π + −
∫ ....(i)
I =
32
6
sin x dx2
π
π
π −
∫
∴∴∴∴ I =
32
6
cos xdx
π
π∫ . ...(ii)
Adding (i) and (ii) we have
∴∴∴∴ 2I = ( )3
2 2
6
sin x cos x dx
π
π∫
∴∴∴∴ 2I =
3
6
dx
π
π∫
Definite Integration
Mahesh Tutorials Science 7
∴∴∴∴ 2I = ( )3
6
x
π
π
∴∴∴∴ 2I = 3 6
π π−
∴∴∴∴ I = 12
π
18)
22
4
cot dx
π
π∫
∴∴∴∴ I = ( )2
2
4
cos ec x 1 dx
π
π
−∫
∴∴∴∴ I =
2 22
4 4
cos ec x dx dx
π π
π π
−∫ ∫
∴∴∴∴ I = ( ) ( )2 2
4 4
cot x x
π π
π π− −
∴∴∴∴ I = cot cot2 4 2 4
π π π π − − − −
∴∴∴∴ I = ( )2
0 14
π − π − − −
I = 14
π−
19)( )
2 2
20
sin xdx
1 cos x
π
+∫
Ans. I = ( )
2 2
20
sin xdx
1 cos x
π
+∫
∴∴∴∴ I =
2
2
220
x x2sin cos
2 2dx
x2cos
2
π
∫
∴∴∴∴ I =
2 22
2 20
x x4sin cos
2 2 dxx x
4cos .cos2 2
π
∫
∴∴∴∴ I =
22
0
xtan dx
2
π
∫
∴∴∴∴ I =
22
0
xsec 1 dx
2
π
−
∫
∴∴∴∴ I =
2 22
0 0
xsec dx dx
2
π π
−∫ ∫
∴∴∴∴ I = ( )2
0
2
0
xtan
2 x1
2
π
π
−
∴∴∴∴ I = tan4 2
π π −
∴∴∴∴ I = 2 – 2
π
GROUP (B)-HOME WORK PROBLEMS
1)
12
0
x 1 + x∫ dx.
Ans. 1 + x2 = t x = 0, t = 1
2x dx = dt x = 1, t = 2
∴∴∴∴ x dx =dt
2
= 1
2
2
1
t dt∫
= 1
2
23/2
1
t
32
= 1
3
23/2
1t
= 1
32 2 1 −
2)
12 5/2
0
x (1 – x )∫ dx
Ans. (1 – x2) = t x = 0, t = 1
x = 1 – t x = 1, t = 0
–2xdx = dt
Definite Integration
8 Mahesh Tutorials Science
∴ I = –1
2
05/2
1
t dt∫
=–1
2
07/2
1
t
72
= –1
7 (0 – 1) =
1
7
3)1
0
1 – x
1 + x∫ dx
Ans. x = cos θ, x = 0, θ = π/2
dx = – sin θ dθ x = 1, θ = 0
= 0
/2
1 – cos
1 + cos π
θ
θ∫ x (– sinθ) dθ
=
0 2
2/2
2sin /2sin
2cos /2d
π
θθ θ
θ× −∫
0
/2
sin /22sin /2 cos /2
cos /2d
π
θθ θ θ
θ= − ×∫
= –2
0
/2
1 – cos d
2π
θθ∫
= ( )0
/2
cos – 1 d
π
θ θ∫
= 0/2
[sin – ]π
θ θ
= – 1 12 2
π π − = −
=2
2
π −
4)
1
2 20
1
(1 + x )∫ dx
Ans. x = tan θ ∴ dx = sec2 θ d θ
∴ When x = 0, θ = 0
x = 1, θ = 4
π
∴∴∴∴ I =
/42
40
1sec d
sec
π
× θ θθ
∫
=
4
20
1
secd
π
θθ
∫
/42
0
cos d
π
θ θ= ∫
=1
2
/4
0
(1 + cos 2 ) d
π
θ θ∫
=4
0
1 sin2
2 2
πθ
θ
+
( )/4
0
12 sin2
4
πθ θ= +
11 0
4 2
π = + −
1
8 4
π= +
1
14 2
π = +
5)
a 2 2
2a/2
a – xdx
x∫
Ans. x = a sin θ, x = a, θ = π/2
dx = a cos θ dθa
2x = , θ = 6
π
∴ I =
/2 2 2 2
2 2/6
a – a sin
a sin
π
π
θ
θ∫ · a cos θ dθ
=
/2
2/6
a cos θ
a sin θ
π
π
∫ cos θ dθ
=
/22
/6
cot d
π
π
θ θ∫
=
/22
/6
(cosec – 1) d
π
π
θ θ∫ .
= /2/6
[–cot ]ππ
θ – /2/6
[ ]ππ
θ
= (0 + 3 ) – – 2 6
π π
= 0 + 3 – 2
6π
= 3 – 3
π
6)
3
20
5x
x + 4∫ dx
Ans. x2 + 4 = t x = 0, t = 4
2xdx = dt x = 3, t = 13
∴∴∴∴ I = 5
2
13
4
1
t∫ dt
= 5
2
13 –12
4
t∫ dt
Definite Integration
Mahesh Tutorials Science 9
= 5
2
131/2
4
t
12
= 5 [ 13 – 2]
7)
1
2 20
dx
(2 – x ) 4 – x∫
Ans. Let, I =
1
2 20
dx
(2 – x ) 4 – x∫
put x = 2 sin θ, dx =2 cos θ dθ,
When x = 0 ∴ θ = 0
when x = 1 ∴ θ =6
π
So,
I = ( )
/6
2 20
2cos
2 4sin 4 4sin
dπ
θ θ
θ θ
− −
∫
( )
/6
2 20
2cos
2 1 2sin 2 1 sin
dπ
θ θ
θ θ
=
− × −∫
( )
/6
0
2cos
2 cos2 2cos
dπ
θ θ
θ θ=
×∫
/6
0
1sec2
2d
π
θ θ= ∫
/6
0
log sec2 tan21
2 2
πθ θ +
=
/6
0
1log sec2 tan2
4
π
θ θ= +
1log 2 3
4= +
8)
ππππ4
2 20
dx
4sin x 5cos x+∫
Ans. Let I =
ππππ4
2 20
dx
4sin x 5cos x+∫
Dividing Nr and Dr by cos2x,
∴ I =
2/4
20
sec x dx
4tan x 5
π
+∫
put tan x = t ∴ sec2 x dx = dt
when x = 0 then t = 0
when x = 4
π then t = 1
So,
I =
1
20
dt
4 5t +∫
( ) ( )
1
220
dt
2 5t
=
+∫
11
0
1 2 1tan
25 5
t− = ×
( )1 11 2tan tan 0
2 5 5
− − = −
11 2
tan2 5 5
− =
9)
/22 2
0
sin cos · cos d
π
θ θ θ θ∫
Ans.
/22 2
0
sin (1 – sin ) · cos d
π
θ θ θ θ∫
put sinθ = t, θ = 0, t = 0
cosθ dθ = dt, θ = 2
π, t = 1
∴∴∴∴ I =
12 2
0
t (1 – t ) dt∫
=
13 5
0
t t –
3 5
= 1
3 –
1
5 =
2
15
BOARD PROBLEMS
1)
22
0
sin x cos x dx
π
∫
Ans. Let sinx = z
cos x dx = dz
When x = 2
π, z = 1
x = 0 , z = 0
Definite Integration
10 Mahesh Tutorials Science
∴∴∴∴ I =
12
0
z dz∫
∴∴∴∴ I =
13
0
z
3
=1
3
2)
24 3
0
sin x cos x dx
π
∫
Ans. I =
24 3
0
sin x cos x dx
π
∫
= ( )2
4 2
0
sin x 1 sin x .cos x dx
π
−∫
Let sinx = z
cos x dx = dz
When x = 2
π, z = 1
x = 0 , z = 0
∴∴∴∴ I = ( )1
4 2
0
z 1 z dz−∫
∴∴∴∴ I = ( )1
4 6
0
z z dz−∫
∴∴∴∴ I =
1 14 6
0 0
z dz z dz−∫ ∫
∴∴∴∴ I =
1 15 7
0 0
z z
5 7
−
∴∴∴∴ I = 1 1
5 7−
∴∴∴∴ I = 2
35
3)
23
0
cos x dx
π
∫
Ans. I =
23
0
cos x dx
π
∫
∴∴∴∴ I = ( )2
0
13cos x cos3x dx
4
π
+∫
∴∴∴∴ I =
2 2
0 0
13 cos x dx cos3xdx
4
π π
+
∫ ∫
∴∴∴∴ I = ( )0
22
0
1 sin3x3 sin x
4 3
ππ
+
∴∴∴∴ I = ( ) ( )1 1
3 1 0 14 3
− + −
∴∴∴∴ I = 1 1
34 3
−
∴∴∴∴ I = 1 8
4 3
=2
3
4)
2
0
cosxdx
1 sinx
π
+∫
Ans. I =
2
0
cosxdx
1 sinx
π
+∫
let 1 + sinx = z
cos x dx = dz
When x = 2
π, z = 2
x = 0 , z = 1
∴∴∴∴ I =
2
1
dz
z∫
∴∴∴∴ I = 2
log z
∴∴∴∴ I = log 2
Definite Integration
Mahesh Tutorials Science 11
5)
2
0
sinxdx
1 cosx
π
+∫
Ans. Let 1 + cos x = z
– sin x dx = dz
∴∴∴∴ sin x dx = – dz
When x = 2
π, z = 1
x = 0 , z = 2
∴∴∴∴ I =
1
2
dz
z−∫
∴∴∴∴ I = 1
2log z−
∴∴∴∴ I = [ ]log1 log 2− −
∴∴∴∴ I = log 2
6)( )
2
30
cos x dx
1 sin x
π
+∫
Ans. Let 1 + sin x = z
cos x dx = dz
When x = 2
π, z = 2
x = 0 , z = 1
∴∴∴∴ I =
2
31
dz
z∫
∴∴∴∴ I =
22
1
z
2
− −
∴∴∴∴ I =
2
21
1 1
2 z
−
∴∴∴∴ I = 1 1
12 4
− −
∴∴∴∴ I = 1 3
2 4
− −
∴∴∴∴ I = 3
8
7)( )
2
30
sinxdx
1 cosx
π
+∫
Ans.
Let 1 + cos x = z
sin x dx = dz
When x = 2
π, z = 1
x = 0 , z = 2
∴∴∴∴ I =
1
32
dz
z
−∫
∴∴∴∴ I =
1-3
2
z dz−∫
∴∴∴∴ I =
1
22
1
2z
−
−
∴∴∴∴ I = 1 1
12 4
−
∴∴∴∴ I = 1 3
2 4
=3
8
8)( ) ( )
2
0
cos x dx
1 sin x 2 sin x
π
+ +∫
Ans. I = ( ) ( )
2
0
cos x dx
1 sin x 2 sin x
π
+ +∫
∴∴∴∴ I = ( ) ( )
2
0
cos x dx
1 sin x 1 sin x 1
π
+ + + +∫
Let 1 + sin x = z
cos x dx = dz
When x = 2
π, z = 2
x = 0 , z = 1
∴∴∴∴ I = ( )
2
1
dz
z z 1+∫
=
2
1
1 1dz
z z 1
−
+ ∫
=
2 2
1 1
1 dzdz
z z 1−
+∫ ∫
=2 2
1 1log z log z 1 − + s
= log 2 – (log 3 – log 2)
= 2 log 2 – log 3
= log 4
3
Definite Integration
12 Mahesh Tutorials Science
9) 20
sin x dx
1 cos x
π
+∫
Ans. Let cos x = z
sin x dx = dz
When x = 2
π, z = 0
x = 0 , z = 1
∴∴∴∴ I =
0
21
dz
1 z
−
+∫
∴∴∴∴ I = ( )0
1
1tan z− −
∴∴∴∴ I = 04
π − −
∴∴∴∴ I = 4
π
10)
2
20
dx
x2cos
2
π
∫
Ans. I =
2
20
dx
x2cos
2
π
∫
∴∴∴∴ I =
22
0
1 xsec dx
2 2
π
∫
∴∴∴∴ I =
2
0
xtan
1 2.12
2
π
∴∴∴∴ I = 2
0
1 x2. . tan2 2
π
∴∴∴∴ I = tan tan04
π−
∴∴∴∴ I = 1
11)
2
0
dx
5 4cos x
π
+∫
Ans. I =
2
0
dx
5 4cos x
π
+∫
∴∴∴∴
2
20
2
dxI
x1 tan
25 4x
1 tan2
π
=
− +
+
∫
∴∴∴∴
22
2 20
x1 tan dx
2I
x x5 1 tan 4 1 tan
2 2
π +
=
+ + −
∫
∴∴∴∴
22
20
xsec dx
2Ix
tan 92
π
=
+∫
Let tanx
2 = z
sec2x
2 dx = 2 dz
When x = 2
π, z = 1
x = 0 , z = 0
∴∴∴∴ I = ( )
1
220
2dz
z 3+∫
∴∴∴∴ I =
1
2 20
dz2
z 3+∫
∴∴∴∴ I =
11
0
1 z2. tan3 3
−
∴∴∴∴ I = 12 1
tan3 3
−
12)
1 -1
20
tan xdx
1 x+∫
Ans. Let tan–1 x = z
2
1dx dz
1 x=
+
When x =1 , z = 2
π
x = 0 , z = 0
∴∴∴∴ I =
2
0
z dz
π
∫
Definite Integration
Mahesh Tutorials Science 13
∴∴∴∴ I = ( )2 4
0
1z
2
π
I = 2
32
π
13)
12 x
20
edx
x∫
Ans. I =
12 x
20
edx
x∫
Let 1
zx
=
∴∴∴∴ 2
1dx dz
x= −
When x = 2 , z = 1
2
x = 0 , z = ∞
∴∴∴∴ I =
12
ze dz
∞
∫
∴∴∴∴ I = ( )1
z 2e∞
−
∴∴∴∴ I = ( )e 0−
∴∴∴∴ I = e−
14)
12
0
1 x dx−∫
Ans. Let x = sin θ
∴∴∴∴ dx = cos θ dθ
x = 1 , θ = 2
π
x = 0 , θ = 0
∴∴∴∴ I =
2
0
cos .cos d
π
θ θ θ∫
I =
2
0
cos .cos d
π
θ θ θ∫
∴∴∴∴ I =
22
0
cos d
π
θ θ∫
∴∴∴∴ I = ( )2
0
1 cos2 d
π
+ θ θ∫
∴∴∴∴ I =
2 2
0 0
1d cos2 d
2
π π
θ + θ θ
∫ ∫
∴∴∴∴ I = ( )2
0
2
0
1 sin2
2 2
π π θ θ +
∴∴∴∴ I =
2
0
1 1 sin2
2 2 2 2
π π θ +
∴∴∴∴ I = ( )1 1
sin 02 2 2
π + π −
∴∴∴∴ I = 1
2 2
π
∴∴∴∴ I = 2
4
π
15) ( )
3
32
dx
x x 1−∫
Ans. I = ( )
3
32
dx
x x 1−∫
∴∴∴∴ I =
( )( )
3 33
32
x x 1dx
x x 1
− −
−∫
∴∴∴∴ I =
3 32
32 2
x 1dx dx
xx 1−
−∫ ∫
I = I1 – I
2
I1 =
3 2
32
xdx
x 1−∫
Let x3 – 1 = z,
3x2 dx = dz
x2 dx = dz
3,
where x = 3, z = 26
x = 2 z = 7
∴∴∴∴ I1 =
26
7
1 dz
3 z∫
Definite Integration
14 Mahesh Tutorials Science
∴∴∴∴ I1 =
26
7
1log z
3
∴∴∴∴ I1 = [ ]1log 26 log 7
3−
=1 26log
3 7
∴∴∴∴ I2 =
3
2
1dx
x∫
∴∴∴∴ I2 =
3
2log x
∴∴∴∴ I2 = log 3 – log 2
∴∴∴∴ I2 =log
3
2
∴∴∴∴ I2 = I
1 – I
2
I = 1 26 3log log
3 7 2−
I =1 26 1 3log .3 log
3 7 3 2−
I =
31 26 3
log log3 7 2
−
I = 1 26 27
log log3 7 8
−
I = 1 26 8log
3 7 27
×
I = 1 208log
3 189
GROUP (C)-HOME WORK PROBLEMS
1)x
0
sin 2x e dx
π
∫
Ans. Let I = x
0
sin 2x e dx
π
∫
/2
x
0I sin2x e dx
π = ∫
( )x
0
de dx sin2x dx
dx
π
− × ∫ ∫
x x
0 0I sin2x.e e cos2x 2 dx
ππ = − × ∫
0 x
0I sin2 .e sin0. 2 cos2x. e dxe
πππ = − −
∫
x
0
I 0 2 cos2x.e dx
π
= − ∫
( )x x
0
I 2 cos2x .e 2sin2x .e dx
ππ
= − − −
∫
0 x
0
I 2 cos2 .e cos0.e 4 sin2x e dx
π
ππ = − − − ∫
I + 4I = – 2 [eπ – 1]
5I = 2(1 – eπ)
∴∴∴∴ I =2
5 (1 – eπ)
2)
/22x
0
e cos x dx
π
∫
Ans. Let, I =
/22x
0
e cos x dx
π
∫
/2
2x
0cos x. dxe
π= ∫
/2
2x
0cos x e dx
π = ∫
( )/2
2x
0
ddx cos x
dxe
π − ∫ ∫ dx
2x/2/22x
0 0
1cos x. sin x
2 2
ee
ππ = − × −
∫ ∫ dx
/2/2
2x 2x
0 0
1 1cos x. sin x.e dx
2 2e
ππ = + ∫
0
2x/2 /22x
0 0
1cos cos0.
2 2
1sin dx cos x dx
2 2
e e
ex e
π
π π
π = − +
−
∫ ∫
∴∴∴∴ I = 1
2[0 – 1] +
/22x
/22x0
0
sin x.1 1
cos xdx2 2 4
e
e
π
π
− ∫
Definite Integration
Mahesh Tutorials Science 15
∴∴∴∴ I + 1
4 I =
01 1sin sin0.
2 4 2e eππ−
+ −
∴∴∴∴5I
4 =
1
2
−+1
4eπ
∴∴∴∴5I 2
4 4
eπ− +
=
∴∴∴∴ I = ( )1
25eπ
−
3) I =log x
42
1
x∫ dx –
4 2
1
x
x∫ dx
Ans. I =
42
1
x∫ log x dx
442 2
11
1log x x dx . x dx dx
x
= − ∫ ∫ ∫
4 43 3
1 1
x 1 x= logx . . dx
3 x 3
−
∫
=
42
1
64 1 1log 4 .0 x dx
3 3 3
− −
∫
43
1
64 1 xlog 4
3 3 3
= −
( )64 1
log 4 64 13 9
= − −
64log 4 7
3= −
4)
43
0
sec x dx∫ππππ
Ans. Let, I =
43
0
sec x dx∫ππππ
/4 2
0sec x sec x dx
π= ∫
/4 2 2
01 tan x sec x dx
π= +∫
put tan x = t, ∴∴∴∴ sec2 x dx = dt
when x = 0 then t = 0
when x = 4
π then t = 1
So,
I = 1 2
01 t+∫ dt
12 2
0
11 log t 1 t
2 2
tt
= × + + + +
[ ]1 1
2 log 1 2 02 2
= + + −
( )1 1
log 1 222
= + +
5)
e2
1
(log x) 1dx∫
Ans. = (logx)2
e
1
1dx∫ –
e
1
12 log x .x dx
x∫
= ( )ee
2
1 1
1x log x 2 log x .x .x dx
x
− − ∫
= e
2
1x (log x)
– 2 ( ) [ ]ee
1 1x log x – x
= e – 2e + 2e –2
= e – 2
6)
13 –1
0
x tan x dx∫
Ans.
1–1 3
0
I = tan x x dx −∫
1–1 3
0
d (tan x) x dx dx
dx
∫ ∫
=
1
0
4 –1[x tan x]
4 –
1
4
1 4
20
x
1 + x∫
1 4
20
1 1 x 1 1dx
4 4 4 1 x
π − += × −
+∫
∴
12
20
1 1I x – 1 +
16 4 x + 1
π= −
∫ dx
[ ]
13
11 100 0
x1
x tan x16 4 3
−
π = − − +
= 16
π –
1
4 1 – 1 +
3 4
π
Definite Integration
16 Mahesh Tutorials Science
= 16
π –
1
12 +
1
4 16
π−
= 1
6
7)
1 –1
2 20
tan xx dx
(1 + x ) 1 + x∫
Ans. put tan–1x = t
2
1
1 + x dx = dt
∴∴∴∴ x = tan t
when x = 0, t = 0
x = 1, t = 4
π
=
/4
20
t tan t dt
1 + tan t
π
∫
=
/4
0
t sin t
cos t1
cos t
π
∫ dt
=
/4
0
tsin t dt
π
∫
= /4
0t sin t
π ∫ –
/4
0
sint dt dt
π
∫ ∫
= [ ]/4
0– t cos t
πs +
/4
0
cos t dt
π
∫
= [ ]/4
0– t cos t
π + [ ]
/40
sin tπ
= –
4 2
π +
1
2
= 1
2 1 –
4
π
8)
12 1
20
xsin xdx
1 x
−
−∫
Ans. Let, sin–1x = π, when x =1
2,
4
ππ =
∴∴∴∴ sin π = x x = 0, π = 0
∴∴∴∴ 2
1dx d
1 x
= π
−
∴∴∴∴ I =
/4
0
sin .d
π
π π π∫
( )/4/4
0 0sin .d 1. sin d d
ππ = π π π − π π π ∫ ∫ ∫
( )/4
0
/4
0cos cos d
π π = − π π − − π π ∫
( )/4
0
1. sin
4 2
π π= − + π
1
4 2 2
π= − +
11
42
π = −
9)
1/ 2 2 –1
20
x sin x dx
1 – x∫
Ans. Let t = sin–1x, x = sint
dt = 2
1 dx
1 – x
when x = 0, t = 0
x = 1
2, 4
π = t
So,
I =
/42
0
sin t · t · dt
π
∫
I = ( )/4
2
0
t sin t dt
π
∫
( )
/4
0
t 1 cos2tdt
2
π−
= ∫
/4 /4
0 0
1t t cos2t dt
2
π π = − ∫ ∫
/4 /42
00
1 t sin2t sin2tt 1. dt
2 2 2 2
π π = − − ∫
/42
0
1 1 cos2tt sin2t
2 32 2 2
π π = − +
/42
0
1 cos2tt sin2t
4 16 2
π π = − +
Definite Integration
Mahesh Tutorials Science 17
21 1 1.1 .0 0
4 16 4 2 2
π π = − + − −
21 1
4 16 4 2
π π= − +
=2 1
64 16 8
π π− +
10)
11
20
xtan dx
1 x
−
− ∫
Ans. Let, I =
11
20
xtan dx
1 x
−
− ∫
put x = sinθ
∴∴∴∴ dx = cosθ dθ
when x = 0 then θ = 0
when x = 1 then θ = π/2
So,
I = /2 -1
0
sinθtan cosθ dθ
cosθ
π
∫
( )/2 1
0tan tanθ cosθ dθ
π −= ∫
/2
0θcosθ dθ
π= ∫
( )/2/2
0 0θ cos θdθ 1. cos θdθ dθ
ππ = − ∫ ∫ ∫
( )/2
0
/2
0θ.sinθ sinθdθ
π π = − ∫
( )/2
0.1 0 cosθ
2
ππ = − − −
( )/2
0cosθ
2
ππ= +
( )0 12
π= + −
12
π= −
BOARD PROBLEMS
1)
1x
0
x e dx∫
Ans. I =
1x
0
x e dx∫
= { }11
x x
0 0x e dx 1. e dx dx
∫ ∫
= ( )1 1
x x
0 0xe e dx −
∫
= ( ) ( )1 1
x x
0 0xe e−
= e – (e – e0)
= e – e + 1
= 1
2)
2x
0
x e dx∫
Ans. I =
2x
0
x e dx∫
= { }22 xx
0 0
x e dx 1. e dx dx −
∫ ∫
=
22 xx
0 0
x e e dx − ∫
= ( ) ( )2 2
x x
0 0x e e−
= 2e2 – (e2 – e0)
= 2e2 – e2 + 1
= e2 + 1
3)
2
0
x sin x dx
π
∫
Ans. I =
2
0
x sin x dx
π
∫
∴∴∴∴ I = ( ) 22
0 0x sinx dx 1 sin x dx dx
ππ − ∫ ∫ ∫
∴∴∴∴ I = ( )2
20
0
x cos x cos x dx
ππ
− − − ∫
∴∴∴∴ I = ( )2
0
2
0
x cos x cos x dx
ππ
− − −∫
∴∴∴∴ I = ( )2
0
2
0
x cos x cos x dx
ππ
− + ∫
Definite Integration
18 Mahesh Tutorials Science
∴∴∴∴ I = ( )2
0.0 0.1 sin x
2
ππ
− − +
∴∴∴∴ I = 0 + (1 – 0)
∴∴∴∴ I = 1
4)
2
0
x dx
1 cos x
π
+∫
Ans. I =
2
0
x dx
1 cos x
π
+∫ =
2
20
xdx
x2cos
2
π
∫
=
22
0
1 xx sec dx
2 2
π
∫
=
222 2
0 0
1 x xx sec dx 1 sec dx dx
2 2 2
ππ −
∫ ∫
=
2 2
0 0
x xtan tan
1 2 2x dx1 12
2 2
π π
−
∫
=
2 2
0 0
1 x x2 x tan 2 tan dx
2 2 2
π π −
∫
∴∴∴∴ I =
2
2
0
0
xlog sec
1 x 22 x tan 2
12 2
2
π
π
−
= 1 0 2 log sec log sec02 4
π π − − − −
= 2.log 22
π−
=1
2. log 22 2
π−
= log 22
π− .
5)
2
0
x log x dx∫
Ans. I =
2
0
x log x dx∫
∴∴∴∴ I =
21
00
1log x. x dx . x dx dx
x
− ∫ ∫ ∫
∴∴∴∴ I =
222 2
0 0
x 1 xlog x. . dx
2 x 2
− ∫
∴∴∴∴ I =
2 22
00
x 1log x xdx
2 2
−
∫
∴∴∴∴ I =
22
0
1 x2log 2 .
2 2
−
∴∴∴∴ I = ( )1
2log 2 44
−
∴∴∴∴ I = 2 log 2 – 1
6)
2
1
log x dx∫
Ans. I =
2
1
log x dx∫
=
22
11
1log x dx . dx dx
x
− ∫ ∫ ∫
= ( )2
1
2
1
1x log x .x dx
x
− ∫
= ( ) ( )2
12log 2 1log1 dx− − ∫
=2 log 2 – (x)2
=2 log 2 – (2–1)
=2 log 2 – 1
7)
e
1
log x dx∫
Ans. =
e
1
log x .1dx∫
=
ee
11
1log x dx . dx dx
x
− ∫ ∫ ∫
Definite Integration
Mahesh Tutorials Science 19
= ( )e
e1
1
1x log x .x dx
x
−
∫
= ( )e
1e log e dx− ∫
= ( )e1
e x−
= e – (e – 1)
= e – e + 1 = 1
8)
11
0
x tan x dx−∫
Ans. =
11
120
0
1tan x x dx . x dx dx
1 x
− − + ∫ ∫ ∫
= ( )
11
2 21
20
0
x xtan x.
2 2 1 x
−
− +
∫
= ( )1
212 1
200
1 1 x 1 1x tan x dx
2 2 1 x
− + −
− +
∫
=
1
20
1 1 dx1. dx
2 4 2 1 x
π − −
+ ∫ ∫
=1
1
0
1x tan x
8 2
−π − −
=11
8 2 4
π π − −
=1
8 2 8
π π− +
=1
4 2
π−
GROUP (D)-HOME WORK PROBLEMS
1)
22
0
x 2 x dx−∫
Ans. Replace x by 2 – x
∴∴∴∴ I = ( ) ( )2
2
0
2 x 2 2 x dx− − −∫
= ( )2 1
2 2
0
4 4x x x dx− +∫
=
2 3 512 2 2
0
4x 4 x dx
− + ∫
=
23 52 2
0
7x4x 4x 23 5 7
2 2 2
− +
3 5 72 2 28 8 2
2 2 22 5 7
− +
=8 8 2
2 2 4 2 8 22 5 7 − +
=16 2 32 2 16 2
3 5 7− +
=560 2 672 2 240 2
105
− +
=128 2
105
2) ( )3a 2 2
0x a x dx−∫
Ans. Replace x by a – x
I = ( ) ( )3a 2 2
0a x a a x dx− − − ∫
= ( )3
a 22
0a x x dx−∫
= ( )3a 2 2 2
0a 2ax x x dx− +∫
=
3 5 7a 2 2 2 20
a x 2ax x dx
− +
∫
=
a5 972 2 2
2
0
x 2ax xa
5 7 9
2 2 2
− +
=5 97
2 2 2 22 4 2a a a a a
5 7 9− +
=92 2 4 2
a5 7 9
− +
=92 14 20 2
a35 9
− +
=92 2 6
a9 35
−
Definite Integration
20 Mahesh Tutorials Science
=92 70 54
a315
−
9216a
315=
3) 20
dx
1 cot x
π
+∫
Ans. I = 20
dx
1 cot x
π
+∫
20
dxI
cot x1
sin x
π
=
+
∫
20
sin xI
sin x cos x
π
=+
∫ dx .... (i)
Replace x by x2
π−
20
sin x2
I dx
sin x cos x2 2
π
π −
=
π π − + −
∫
20
cos xI
cos x sin x
π=
+∫ dx ....(ii)
Adding (i) and (ii)
∴∴∴∴2
0
sin x cos x2I dx
sin x cos x
π += + ∫
∴∴∴∴ 20
2I dxπ
= ∫
∴∴∴∴ [ ]2
02I x
π
=
∴∴∴∴ 2I = 2
π
∴∴∴∴ I = 4
π
4)( )
2
20
sin d
sin cos
π
θ θ
θ + θ∫
Ans. I = ( )
2
20
sin d
sin cos
π
θ θ
θ + θ∫ ....(i)
2
20
sin d2
I
sin cos2 2
π π − θ θ
=
π π − θ + − θ
∫
( )
2
20
cos dI
cos sin
π
θ θ=
θ + θ∫ ...(ii)
Adding (i) and (ii)
∴∴∴∴( )
( )
2
20
sin cos d2I
sin cos
π
θ + θ θ=
θ + θ∫
∴∴∴∴
2
0
1d2I
sin cos
π
θ=
θ + θ∫
∴∴∴∴
2
0
1d2I
1sin 1cos
π
θ=
θ + θ∫
1 = r cos α ; 1 = r sin α
2 2r a b= +
2 2r 1 1= +
r 2=
tan α = 1
α =4
π
20
d2I
r cos sin r sin cos
π θ=
α θ + α θ∫
[ ]2
0
d2I
r sin cos cos sin
π θ=
α θ + α θ∫
[ ]2
0
1 d2I
rsin
π θ=
α + θ∫ ( )/2
0
1cos dec
r
πα θ θ= +∫
( ) ( )2
0
12I log cosec cot
r
π
= α + θ − α + θ
( ) ( )
12I log cosec cot
2 22
log cosec 0 cot 0
π π = α + − + α −
+ α − + α
1
2I log sec tan log cosec cot2
= α + α − α − α
1
2I log sec tan log cos ec cot4 4 4 42
π π π π = + − −
Definite Integration
Mahesh Tutorials Science 21
1
2I log 2 1 log 2 12 = + − +
1 2 1
2I log2 2 1
+=
−
( ) ( )
( ) ( )
2 1 2 112I log
2 2 1 2 1
+ +=
− +
( )2
2 112I log
12
+=
12I 2log 2 1
2= +
∴∴∴∴ I = 1log 2 1
2+
5)
3
0 2
dx
x 9 x+ −∫
Ans. I=
3
0 2
dx
x 9 x+ −∫
x = 3 sin θ
dx = 3 cosθ dθ
x = 3 sinθ
When x = 3; x = 0
3 = 3 sinθ ; 0 = 3sinθ
1 = sinθ ; 0 = sinθ
2
πθ = ∴∴∴∴ θ = 0
/2
0 2
3cos dI
3sin 9 9sin
π θ θ=
θ + − θ∫
/2
0
3cos dI
3sin 3cos
π θ θ=
θ + θ∫
( )
/2
0
3cos dI
3 sin cos
π θ θ=
θ + θ∫
/2
0
cos dI
sin cos
π θ θ=
θ + θ∫ ....(i)
Replace θ by 2
π− θ
/2
0
cos d2
I
sin cos2 2
π
π − θ θ
=
π π − θ + − θ
∫
/2
0
sin dI
cos sin
π θ θ=
θ + θ∫ ....(ii)
Adding (i) and (ii)
∴∴∴∴ ( )
( )
/2
0
cos sin d2I
cos sin
π θ + θ θ=
θ + θ∫
[ ] 20
2Iπ
= θ
2I = 02
π−
2I = 2
π
∴∴∴∴ I = 4
π
6) ( )20
sin2x log tan x dxπ
∫
Ans. I = ( )20
sin2x log tan x dxπ
∫ ....(i)
Replace x by x2
π−
20
I sin2 x log tan x dx2 2
π π π = − −
∫
( ) ( )20
I sin 2x log cot x dxπ
= π −∫
( )20
I sin2x log cot x dxπ
= ∫ ....(ii)
Adding (i) and (ii)
∴∴∴∴ ( ) ( )20
2I sin2x log tan x log cot x dxπ
= + ∫
∴∴∴∴ ( )20
2I sin2x log tan x cot x dxπ
= ∫
∴∴∴∴ ( )20
2I sin2x log 1 dxπ
= ∫
∴∴∴∴ 2I = 0 [ ]log1 0=∵
∴∴∴∴ I = 0
BOARD PROBLEMS
1)
a
0
xdx
x a x+ −∫
Ans. I =
a
0
xdx
x a x+ −∫ .....(i)
Definite Integration
22 Mahesh Tutorials Science
I = ( )
a
0
a xdx
a x a a x
−
− + − −∫
( ) ( )a a
0 0
f x dx f a x dx = − ∫ ∫∵
I =
a
0
a xdx
a x a a x
−
− + − +∫
I =
a
0
a xdx
a x x
−
− +∫
I =
a
0
a x
x a x
−
+ −∫ .....(ii)
Adding (i) and (ii), we have
2I =
a
0
x a xdx
x a x x a x
−+ + − + −
∫
∴∴∴∴ 2I =
a
0
dx∫
∴∴∴∴ I = ( )a0
x
∴∴∴∴ I = ( )1
a2
∴∴∴∴ I = a
2
2) ( )1
32 2
0
x 1 x dx−∫
Ans. I = ( )1
32 2
0
x 1 x dx−∫
I = ( ) ( )1 32 2
0
1 x 1 1 x dx− − − ∫
( ) ( )a a
0 0
f x dx f a x dx = − ∫ ∫∵
I = ( ) ( )1
32 2
0
1 2x x 1 1 x dx− + − +∫
I = ( )1 3
2 2
0
1 2x x x dx− +∫
I =
1 3 5 72 2 2
0
x 2x x dx
− + ∫
I =
1 1 13 5 72 2 2
0 0 0
x dx 2 x dx x dx− +∫ ∫ ∫
I =
1 115 972 2 2
00 0
2 2 2x 2 x x
5 7 9
− × +
I = 2 4 2
5 7 9− +
I =126 180 70
315
− +
I = 196 180
315
−
I = 16
315
3)
1
20
dx
x 1 x+ −∫
Let x = sinθ
∴∴∴∴ dx = cosθ dθ
When x = 1 θ =2
π
x = 0 θ = 0
I =
2
0
cos d
sin cos
π
θ θ
θ + θ∫ ....(i)
I =
2
0
cos d2
sin cos2 2
π π − θ θ
π π
− θ + − θ
∫
( ) ( )a a
0 0
f x dx f a x dx = − ∫ ∫
I =
2
0
sin d
cos sin
π
θ θ
θ + θ∫ ....(ii)
Adding (i) and (ii)
2I =
2
0
cos sind
sin cos cos sin
π
θ θ + θ
θ + θ θ + θ ∫
∴∴∴∴ 2I =
2
0
sin cosd
sin cos
π
θ + θ θ
θ + θ ∫
Definite Integration
Mahesh Tutorials Science 23
∴∴∴∴ 2I =
2
0
d
π
θ∫
∴∴∴∴ 2I = ( )2
0
π
θ
∴∴∴∴ 2I = 2
π
∴∴∴∴ I = 4
π
4)
22
0
cos x dx
π
∫
Ans. I =
22
0
cos x dx
π
∫ ....(i)
I =
22
0
cos x dx2
π
π −
∫
( ) ( )a a
0 0
f x dx f a x dx = − ∫ ∫∵
I =
22
0
sin x dx
π
∫ ....(ii)
Adding (i) and (ii), we have
2I = ( )2
2 2
0
sin x cos x dx
π
+∫
∴∴∴∴ 2I =
2
0
dx
π
∫
∴∴∴∴ 2I = ( )2
0x
π
∴∴∴∴ 2I = 2
π
∴∴∴∴ I = 4
π
5)
22
0
sin x dx
π
∫
Ans. I =
22
0
sin x dx
π
∫ ....(i)
I =
22
0
sin x dx2
π
π −
∫
( ) ( )a a
0 0
f x dx f a x dx = − ∫ ∫∵
I =
22
0
cos x dx
π
∫ ...(ii)
Adding (i) and (ii) we have
2I= ( )2
2 2
0
sin x cos x dx
π
+∫
2I =
2
0
dx
π
∫
2I = ( )2
0x
π
2I = 2
π
I = 4
π
6)
2
0
cos x dx
sin x cos x
π
+∫ ...(i)
Ans. I =
2
0
cos x2
dx
sin x cos x2 2
π π −
π π
− + −
∫
( ) ( )a a
0 0
f x dx f a x dx = − ∫ ∫∵
I =
2
0
sin x dx
cos x sin x
π
+∫ ...(ii)
Definite Integration
24 Mahesh Tutorials Science
Adding (i) and (ii), we have
2I =
2
0
sin x cos xdx
cos x sin x sin x cos x
π
+
+ + ∫
2I =
2
0
dx
π
∫
2I = ( )2
0x
π
2I = 2
π
I = 4
π
7)
2
0
dx
1 tan x
π
+∫
Ans. I =
2
0
dx
1 tan x
π
+∫
I =
2
0
dx
sin x1
cos x
π
+∫
I=
2
0
cos xdx
cos x sinx
π
+∫ ....(i)
I =
2
0
cos x2
dx
cos x sin x2 2
π π −
π π
− + −
∫
( ) ( )a a
0 0
f x dx f a x dx = − ∫ ∫∵
I =
2
0
sin xdx
sin x cos x
π
+∫
Adding (i) and (ii), we have
2I =
2
0
sin x cos xdx
cos x sin x sin x cos x
π
+
+ + ∫
2I =
2
0
dx
π
∫
2I = ( )2
0x
π
2I = 2
π
I = 4
π
8)
2
0
dx
1 cot x
π
+∫
Ans. I =
2
0
dx
1 cot x
π
+∫
I =
2
0
dx
cos x1
sin x
π
+∫
I=
2
0
sin xdx
sin x cos x
π
+∫ ...(i)
I =
2
0
sin x2
dx
sin x cos x2 2
π π −
π π
− + −
∫
( ) ( )a a
0 0
f x dx f a x dx = − ∫ ∫∵
I =
2
0
cos x dx
cos x sin x
π
+∫ ...(ii)
Adding (i) and (ii), we have
2I =
2
0
sin x cos xdx
cos x sin x sin x cos x
π
+
+ + ∫
2I =
2
0
dx
π
∫
2I = ( )2
0x
π
2I = 2
π
I = 4
π
Definite Integration
Mahesh Tutorials Science 25
9)
2
0
sin x dx
sin x cos x
π
+∫
Ans. I =
2
0
sin x dx
sin x cos x
π
+∫ ...(i)
I =
2
0
sin x2
dx
sin x cos x2 2
π π −
π π − + −
∫
( ) ( )a a
0 0
f x dx f a x dx = − ∫ ∫∵
I =
2
0
cos xdx
cos x sin x
π
+∫ ...(ii)
Adding (i) and (ii), we have
2I =
2
0
sin x cos xdx
sin x cos x cos x sin x
π
+ + +
∫
2I =
2
0
dx
π
∫
2I = ( )2
0x
π
2I = 2
π
I = 4
π
10)
2
0
sec xdx
sec x cos ecx
π
+∫
Ans. I =
2
0
sec xdx
sec x cos ecx
π
+∫ ...(i)
I =
2
0
sec x2
dx
sec x cos ec x2 2
π π −
π π − + −
∫
( ) ( )a a
0 0
f x dx f a x dx = − ∫ ∫∵
I =
2
0
cos ecxdx
sec x sec x
π
+∫ ...(ii)
Adding (i) and (ii), we have
2I =
2
0
secx cos ecxdx
sec x cos ecx cos ec x sec x
π
+
+ + ∫
2I =
2
0
dx
π
∫
2I = ( )2
0x
π
2I = 2
π
I = 4
π
11)
a
2 20
dx
x a x+ −∫
Ans. I =
a
2 20
dx
x a x+ −∫
Let x = a sin θ ∴∴∴∴ dx = a cos θ dθ
when x = a, θ = 2
π
x = 0, θ = 0
∴∴∴∴ I =
2
0
a cos d
a sin a cos
π
θ θ
θ + θ∫ .
I =
2
0
cos d
sin cos
π
θ θ
θ + θ∫ ...(i)
I =
2
0
cos d2
sin cos2 2
π π − θ θ
π π
− θ + − θ
∫
( ) ( )a a
0 0
f x dx f a x dx = − ∫ ∫∵
∴∴∴∴ I =
2
0
sin d
cos sin
π
θ θ
θ + θ∫ ...(ii)
Adding (i) and (ii), we have
Definite Integration
26 Mahesh Tutorials Science
2I =
2
0
cos sind
sin cos cos sin
π
θ θ + θ
θ + θ θ + θ ∫
∴∴∴∴ 2I =
2
0
d
π
θ∫
∴∴∴∴ 2I = ( )2
0x
π
∴∴∴∴ 2I = 2
π
∴∴∴∴ I = 4
π
12)( )
2 2
20
sin xdx
sin x cos x
π
+∫
Ans. I = ( )
2 2
20
sin xdx
sin x cos x
π
+∫ ...(i)
I =
22
20
sin x2
sin x cos x2 2
π π −
π π − + −
∫
( ) ( )a a
0 0
f x dx f a x dx = − ∫ ∫∵
I=( )
2 2
20
cos xdx
cos x sin x
π
+∫ ...(ii)
Adding (i) and (ii), we have
2I = ( ) ( )
2 2 2
2 20
sin x cos xdx
sin x cos x cos x sin x
π + + +
∫
∴∴∴∴ 2I =( )
2
20
dxdx
sin x cos x
π
+∫
Now
cos x + sin x =1 1
2 cos x. sin x.2 2
+
= 2 cos x cos sin4xsin4 4
π π +
= 2 cos x4
π −
∴∴∴∴ 2I =
2
20
dx
2 cos x4
π
π −
∫
∴∴∴∴ 2I=
22
0
1sec x dx
2 4
π
π −
∫
∴∴∴∴ 2I =2
0
1tan x
2 4
π π
−
∴∴∴∴ 2I =1
tan tan4 2 4 4
π π π − − −
∴∴∴∴ I =1
tan tan4 4 4
π π +
∴∴∴∴ I = 1
24
×
∴∴∴∴ I = 1
2
13) ( )4
0
log 1 tan x dx
π
+∫
Ans. I = ( )4
0
log 1 tan x dx
π
+∫
I =
4
0
log 1 tan x dx4
π
π + −
∫
I =
4
0
tan tanx4log 1 dx
1 tan tanx4
π π −
+ π +
∫
I =
4
0
1 tanxlog 1 dx
1 tanx
π
− + +
∫ .
I =
4
0
1 tan x 1 tan xlog dx
1 tan x
π
+ + − +
∫
Definite Integration
Mahesh Tutorials Science 27
I =
4
0
2log dx
1 tan x
π
+ ∫
∴∴∴∴ I = ( )4
0
log 2 log 1 tanx dx
π
− + ∫
∴∴∴∴ I = ( )4 4
0 0
log 2 dx log 1 tan x dx
π π
+∫ ∫
∴∴∴∴ I = ( )4
0log 2 x
π
...(i)
∴∴∴∴ 2I = log 24
π
∴∴∴∴ I = log 28
π
14)
2 2
0
sin x dx
sin x cos x
π
+∫
Ans. I =
2 2
0
sin x dx
sin x cos x
π
+∫ ...(i)
∴∴∴∴ I =
22
0
sin x2
dx
sin x cos x2 2
π π −
π π
− + −
∫
( ) ( )a a
0 0
f a dx f a x dx = − ∫ ∫∵
I =
2 2
0
cos xdx
cos x sin x
π
+∫ ...(ii)
Adding (i) and (ii) we have,
2I =
2 2 2
0
sin x cos xdx
sin x cos x sin x cos x
π
+ + + ∫
2I =
2
0
dx
sin x cos x
π
+∫
Now
sinx + cos x =
2
2 2
x x2tan 1 tan
2 2x x
1 tan 1 tan2 2
−
+
+ +
=
2
2
x x2tan 1 tan
2 2x
sec2
+ −
∴∴∴∴ 2I =
2
20
2
dx
x x2tan 1 tan
2 2x
sec2
π
+ −∫
∴∴∴∴ 2I =
22
20
xsec dx
2x x
2tan 1 tan2 2
π
+ −∫
Let tan 2
π= z sec2
2
πdx = 2dz
When x = 2
π, z = 1
x = 0, z = 0
2I =
1
20
2dz
2z 1 z+ −∫
= ( )
1
20
dz2
2 z 2z 1− − +∫
∴∴∴∴ 2I = ( ) ( )
1
2 20
dz2
2 z 1− −∫
∴∴∴∴ 2I =
1
0
1 2 z 1log
2 2 2 z 1
+ −
− +
∴∴∴∴ 2I =1 2 2 1
log log2 2 2 2 1
−−
−
∴∴∴∴ I =
( )
( )
2 11 2 1log
2 2 2 12 1
− − − × −+
∴∴∴∴ I =
( )2
2 11log
2 12 2
−
− −
∴∴∴∴ I = ( )1
.2log 2 12 2
− −
∴∴∴∴ I = ( )1log 2 1
2− −
Definite Integration
28 Mahesh Tutorials Science
∴∴∴∴ I = ( )1 1
log1 log 2 12 2
− − −
∴∴∴∴ I = 1 1 2 1
log2 2 1 2 1
+− × − +
∴∴∴∴ I = 1 2 1
log2 2 1
+ +
∴∴∴∴ I = ( )1
log 2 12
+
GROUP (E)-HOME WORK PROBLEMS
1)
1 2
21
1 xdx
1 x−
−
+∫
Ans. I =
1 2
21
1 xdx
1 x−
−
+∫
( )2
2
1 xf x
1 x
−=
+
( )( )( )
2
2
1 xf x
1 x
− −
− =
+ − = ( )
2
2
1 xf x
1 x
−− =
+ = f(x)
∴∴∴∴ f(x) is even function
1 2
20
1 xI 2 dx
1 x
−=
+∫
( )( )21
20
x 1 22 dx
1 x
− + +
=+
∫
1 1
20 0
dxI 2 dx 2
1 x
= − + + ∫ ∫
11
0I 2 x 2tan x− = − +
1I 2 1 2tan 1 0− = − + −
I 2 1 24
π = − + ×
I 2= π −
2)
1 2
21
xdx
x 1− +∫
Ans. f (x) =
2
2
x
x 1+
f (– x) =( )
( )
2 2
2 2
x x
x 1x 1
−=
+− +
f (– x) = f(x)
∴∴∴∴ f (x) is even function
∴∴∴∴
1 2
20
xI 2 dx
x 1=
+∫
∴∴∴∴
1 2
20
x 1 1I 2 dx
x 1
+ −=
+∫
1
20
1I 2 1 dx
x 1
= −
+ ∫
1 10I 2 x tan x− = −
1I 2 1 tan 1− = −
I 2 14
π = −
∴∴∴∴ I 22
π= −
3)
a 3
2a
xdx
4 x− −∫
Ans. I =
a 3
2a
xdx
4 x− −∫
f (x) =
3
2
x
4 x−
f (– x) = ( )
( )
3
2
x
4 x
−
− − =
3
2
x
4 x
−
−
f (– x) = – f (x)
∴∴∴∴ f (x) is odd function
a 3
2a
xI dx
4 x−
=−
∫
I = 0
4)
26
2
cos x dx
π
−π∫
Ans. I =
26
2
cos x dx
π
−π∫
f (x) = cos 6 x
f (– x) = cos6 (– x) = cos 6 x
∴∴∴∴ f (x) is even function
Definite Integration
Mahesh Tutorials Science 29
I =
26
0
2 cos x dx
π
∫
n 1n
n 2
cos x sin x n 1cos x dx
4 n
cos xdx
−
−
−= +
∫
∫
5 32cos x sin x 5 cos x sin x 3
I 2 cos x dx6 6 4 4
= + +
∫
53 2cos x sin x 5 15
I 2 cos x sin x cos x dx6 24 24
= + +
∫
53
10
cos x sin x 5 15cos x sin x
6 24 24I 2
cos x sin x 1cos x
2 2
+ +
=
+
∫
5 23
0
cos x sin x 5 15cos x sin x
6 24 48I 215
cos x sin x x48
π
+ + =
+
15
I 2 0 0 0 048 2
π = + + + −
2 15
I48 2
× π=
×
5
I16
π=
5)
43
4
tan xsec x dx
π
−π∫
f (x) = tan3 x sec x
f (–x) = tan3 (– x) sec (– x)
= – tan3 x sec x
f (– x) = – f (x)
∴∴∴∴ f (x) is odd function
∴∴∴∴ I =
43
4
tan xsec x dx 0
π
−π
=∫
GROUP (F)-HOME WORK PROBLEMS
1) ( )2
0
log cos x dx
π
∫
Ans. I = ( )2
0
log cos x dx
π
∫ ...(i)
Replace x by x2
π−
I =
2
0
log cos x dx2
π
π −
∫
I = ( )2
0
log sin x dx
π
∫ ...(ii)
Adding (i) and (ii)
∴∴∴∴ 2I = ( ) ( )2
0
log sin x log cos x dx
π
+∫
∴∴∴∴ 2I =
2
0
2sin x cos xlog dx
2
π
∫
2I = ( )2
0
log log sin2x log 2 dx
π
−∫
2I =
/22
0 0
log sin2x dx log 2 dx
ππ
−∫ ∫
2I = I1 – I
2 ...(i)
∴∴∴∴ I1 =
0
1log sin d
2
π
π π∫
=
2.2
0
1log sin d
2
π
π π∫
( ) ( )If f 2a x f x i.e.sin 2. sin2
π − = − π = π
/2
0
1.2 log sin d
2
π
= π π∫
( ) ( )/2 b b
0 a a
log sin d f x dx f y dy
π = π π =
∫ ∫ ∫
Definite Integration
30 Mahesh Tutorials Science
( ) ( )/2 a a
0 0 0
log sin x dx f x dx f a x da2
π π = − = −
∫ ∫ ∫
/2
0
log cos xdx
π
= ∫
= I ....(ii)
I2
/2
0
log 2 dx
π
= ∫
= ( )/2
0x log 2
π
...(iii)
= log 22
π
from (i)
2I = log 2 log 22
π (by ii and iii)
∴∴∴∴ I = log 22
π−
∴∴∴∴ I = log 22
π−
∴∴∴∴ I = log1 log 22 2
π π−
=1
log2 2
π
2)
14
0
sin x cos x dx∫
Ans. I =
14
0
sin x cos x dx∫
cos x = t
– sin x dx = dt
sin x dx = – dt
cos x = t
When x = 1 ; x = 0
cos 1 = t ; cos 0 = t
0 = t; 2
π= t
( )0
4
2
I t dt
π
= −∫
24
0
I t dt
π
= ∫
5 2
0
tI
5
π
=
51
I 05 2
π = −
51I
5 32
π= ×
5
I160
π=
3)0
x sin xI dx
1 sin x
π
=+∫
Ans.( ) ( )
( )0
x sin xI dx
1 sin x x
ππ − π −
=+ π −∫
∴∴∴∴( )
0
x sin xI dx
1 sin x
ππ −
=+∫
∴∴∴∴0 0
sin x x sin xI dx dx
1 sin x 1 sin x
π π
= π −+ +∫ ∫
∴∴∴∴ 2I = ( )
( ) ( )0 0
sin x 1 sin xsin x dxdx
1 sin x 1 sin x 1 sin x
π π−
π = π+ + −∫ ∫
∴∴∴∴ 2I = ( )
20
sin x 1 sin xdx
cos x
π−
π∫
∴∴∴∴ 2I =
2
20
sin x sin xdx
cos x
π −π
∫
∴∴∴∴ 2I = ( )2
0
sec x tan x tan x dx
π
π −∫
∴∴∴∴ 2I = 2
0 0
sec x tan x dx tan xdx
π π π − ∫ ∫
∴∴∴∴ 2I = 2
0 0 0
sec x tan x dx sec x dx dx
π π π π − + ∫ ∫ ∫
∴∴∴∴ 2I = ( ) ( ) ( )0 0 0
sec x tan x xπ π π
π − +
∴∴∴∴ 2I = ( ) ( )1 1 0 0π − − − − + π
Definite Integration
Mahesh Tutorials Science 31
∴∴∴∴ 2I = ( )22
π− + π = ( )2
2
ππ −
4)
/2
0
x2
sin x cos x2 2
ππ
−
π π − + −
∫
Ans.
/2
0
x2
I
sin x cos x2 2
ππ
−
=π π
− + −
∫
∴∴∴∴
/2
0
x dx2
Icos x sin x
ππ
−
=+∫
∴∴∴∴
/2 /2
0 0
dx xdxI
2 sin x cos x sin x cos x
π ππ
= −+ +
∫ ∫
∴∴∴∴ 2I =
/2
0
dx
2 sin x cos x
ππ
+∫
sin x + cosx =
2
2 2
x x2tan 1 tan
2 2x x
1 tan 1 tan2 2
−
+
+ +
2
2
x x2tan 1 tan
2 2x
1 tan2
+ −
=
+
∴∴∴∴
2/2
20
x1 tan dx
22I
x x22tan 1 tan
2 2
π
+ π
=
+ −∫
∴∴∴∴
2/2
20
xsec
22I dxx x2
2tan 1 tan2 2
ππ
=
+ −∫
Let tan 2
α= π
2 x 1sec . dx d
2 2
= π
∴∴∴∴2 x
sec . dx 2d2
= π
when x2
π= , π = 1
x = 0 π = 0
∴∴∴∴ 2I =
1
20
2d
2 2 1
π π
π + − π∫
∴∴∴∴ 2I = ( )
1
20
d.2
2 2 2 1
π π
− π − π +∫
∴∴∴∴ 2I = ( ) ( )
1
2 20
d
2 1
ππ
− π −∫
∴∴∴∴ 2I =
1
0
1 2 1. log2 2 2 1
+ π −π
− π +
∴∴∴∴ 2I = 2 2 1
log log2 2 2 2 1
π −− +
∴∴∴∴ 2I =
( ) ( )
( ) ( )
2 1 2 10 log
2 2 2 1 2 1
− −π − + −
∴∴∴∴ 2I = ( )
22 1
log2 12 2
−π
−
∴∴∴∴ 2I = .2log 2 12 2
π−
∴∴∴∴ I = ( )log 2 12 2
π− −
I = ( )log 2 12 2
π+
5) ( )/2
0
log sin x dx
π
∫
Ans. ( )/2
0
I log sin x dx
π
= − ∫ ...(i)
/2
0
log sin x dx2
ππ
= − −
∫
/2
0
log cos x dx
π
= − ∫ ...(ii)
Adding (i) and (ii) we have
2I = ( )/2
0
log sin x log cos x dx
π
− +∫
Definite Integration
32 Mahesh Tutorials Science
∴∴∴∴ 2I =
/2
0
2sin x cos xlog dx
2
π
−
∫
∴∴∴∴ 2I = ( )/2
0
log sin2x log 2 dx
π
− −∫
∴∴∴∴ 2I =
/2 /2
0 0
log sin2x dx log 2 dx
π π
− +∫ ∫
∴∴∴∴ 2I = I1 + I
2
I1 =
/2
0
log sin2x dx
π
− ∫
put 2x = π
dx =d
2
π
when x = 2
π , z = π
x = 0, z = 0
∴∴∴∴ I1 =
0
1log sin z dz
2
π
− ∫
∴∴∴∴ I1 =
0
1log sin x dx
2
π
− ∫
∴∴∴∴ I1 =
2.2
0
1log sin x dx
2
π
− ∫
∴∴∴∴ I1=
2.2
0
1log sin x dx
2
π
− ∫
∴∴∴∴ I1 ( ) ( )
0
1.2 log sin x dx f 2a x f x
2
π
− − = ∫ ∵
∴∴∴∴ I1 = I ...(iii)
I2 = log 2
/2
0
dx
π
∫
∴∴∴∴ I2 = ( )
/2
0x log 2
π
∴∴∴∴ I2 = log 2
2
π .....(iv)
∴∴∴∴ 2I = I + log 22
π(by iii and iv)
∴∴∴∴ I = log 22
π
6)
2
0
1x dxcos xIsin x
1cos x
π
=
+∫
Ans.
2
0
1x dxcos xIsin x
1cos x
π
=
+∫
2
0
xdx
cos xIcos x sin x
cos x
π
=+∫
∴∴∴∴
2
0
x dxI
sinx cos x
π
=+∫
Refer as Group F class work (5)
5) ( )20
log cos ecx dxπ
∫
I = ( )20
log sin x dxπ
−∫
I = ( )20
log sin x dxπ
−∫ ...(i)
Refer as Group F class work (2)
GROUP (G)-HOME WORK PROBLEMS
Q -1)
9 3
3 33
12 xdx
x 12 x
−
+ −∫
Ans. Let I =
9 3
3 33
12 xdx
x 12 x
−
+ −∫ ...(i)
using ( ) ( )b b
a af x dx f a b x dx= + −∫ ∫
=( )
( )
39
33 3
12 12 xdx
12 x 12 12 x
− −
− + − −∫
=
39
333
xdx
12 x x− +∫ ... (ii)
Adding (i) and (ii)
2I =
3 39
3 33
x 12 xdx
x 12 x
+ −
+ + −∫
∴∴∴∴ 2I = [ ]9 9
331dx x=∫
∴∴∴∴ 2I = 6
∴∴∴∴ I = 3
Definite Integration
Mahesh Tutorials Science 33
2)( )
( )
26
22 2
8 xdx
x 8 x
−
+ −∫
Ans. Let I =( )
( )
26
22 2
8 xdx
x 8 x
−
+ −∫ ...(i)
( )
( ) ( )
26
22 2
8 8 xdx
8 x 8 8 x
− − =
− + − − ∫
∴∴∴∴ ( )
26
22 2
xI dx
8 x x=
− +∫ ...(ii)
Adding (i) and (ii)
∴∴∴∴ ( )
( )
2 26
22 2
8 x x2I dx
x 8 x
− +=
+ −∫
∴∴∴∴ [ ]6 6
222I 1dx x= =∫
∴∴∴∴ 2I = 4
∴∴∴∴ I = 2
3)
13
3 31
x 52I dx
x 5 9 x
+=
+ + −∫
Ans. I =
13
3 31
x 52I dx
x 5 9 x
+=
+ + −∫ ...(i)
using ( ) ( )b b
a a
f x dx a b x dx= + −∫ ∫
33
3 31
9 xI dx
9 x 5 x
−=
− + +∫ ...(ii)
Adding (i) and (ii)
∴∴∴∴
3 33
3 31
9 x x 52I dx
9 x x 5
− + +=
− + +∫
∴∴∴∴ [ ]3 3
112I dx x= =∫
∴∴∴∴ 2I = 2
I = 1
4)
3
6
1dx
1 tanx
π
π +∫
Ans. Let I =
3
6
1dx
1 tanx
π
π +∫
3
6
1dx
1 sin x/cos x
π
π
=+
∫
3
6
cos xdx
cos x sin x
π
π
=+
∫ ... (i)
( ) ( )b b
a a
f x dx a b x dx= + −∫ ∫
∴∴∴∴
3
6
cos x2
I dx
cos x sin x2 2
π
π
π −
=
π π − + −
∫
3
6
sin xI dx
cos x sin x
π
π
=+
∫ ... (ii)
Adding (i) and (ii)
∴∴∴∴
3
6
sin x cos x2I dx
sin x cos x
π
π
+=
+∫
∴∴∴∴[ ]
3
6
3
6
2I dx x
π
π
π
π
= =∫
∴∴∴∴ 2I =6
π
I = π / 12
5)3
36 2
dx
1 cot x
π
π
+
∫
Ans. I = 3
36 2
dx
1 cot x
π
π
+
∫ ...(i)
( )
33
6 2
dxdx
1 cos x/sin x
π
π=
+∫
∴∴∴∴
32
33 3
6 2 2
sin xI dx
sin x cos x
π
π=
+
∫ ...(ii)
Adding (i) and (ii)
∴∴∴∴ [ ]3
6
3
6
2I 1dx x6
π
π
π
ππ= = =∫
Definite Integration
34 Mahesh Tutorials Science
∴∴∴∴ I = 12
π
6)
333 3
6
sin xdx
sin x cos x
π
π+
∫
Ans. I =
333 3
6
sin xdx
sin x cos x
π
π+
∫ ...(i)
using ( ) ( )b b
a a
f x dx a b x dx= + −∫ ∫
3
3
6 3 3
sin x2
dx
sin x cos x2 2
π
π
π −
=
π π − + −
∫
∴∴∴∴
333 3
6
cos xI dx
cos x sin x
π
π=
+∫ ... (ii)
Adding (i) and (ii)
∴∴∴∴
3 333 3
6
cos x sin x2I dx
cos x sin x
π
π
+=
+∫
∴∴∴∴ [ ]3
6
3
6
2I 1dx x
π
π
π
π= =∫
∴∴∴∴ 2I = π /6
∴∴∴∴ I = 12
π
7)7
2
xdx
x 9 x+ −∫
Ans. I = 7
2
xdx
x 9 x+ −∫ ... (i)
( )
7
2
9 xdx
9 x 9 9 x
−=
− + − −∫
7
2
9 xdx
9 x x
−=
− +∫ ... (ii)
Adding (i) and (ii)
∴∴∴∴7
2
9 x x2I dx
9 x x
− +=
− +∫
∴∴∴∴ [ ]7 7
222I 1. dx x= =∫
∴∴∴∴ 2I = 5
∴∴∴∴ I = 5/2
8)5
2
xdx
7 x x− +∫
Ans. I = 5
2
xdx
7 x x− +∫ ... (i)
I = ( )
5
2
7 xdx
7 7 x 7 x
−
− − + −∫
5
2
7 xdx
x 7 x
−
+ −∫ ... (ii)
Adding (i) and (ii)
∴∴∴∴ 2I = 5
2
7 x xdx
7 x x
− +
− +∫
∴∴∴∴ 2I = [ ]5 5
221 dx x=∫
∴∴∴∴ 2I = 3
∴∴∴∴ I = 3/2
GROUP (H)-HOME WORK PROBLEMS
1) ( )5
24x 3 dx+∫
Ans. Let I = ( )5
24x 3 dx+∫
Let f(x) = 4x + 3; a = 2, b = 5
h = b a
n
−= 5 2
n
−=3
n
∴∴∴∴ nh = 3
∴∴∴∴ ( ) ( )5
21
lim f a rhn
n r
f x dx h→∞
=
= +∑∫
= ( )1
lim f 2 rhn
n r
h→∞
=
+∑
( )( )1
lim 4 2 rh 3n
n r
h→∞
=
= + +∑
( )1
lim 11 4rhn
n r
h→∞
=
= +∑
2
1 1
lim 11 4n n
n r r
h h r→∞
= =
= +
∑ ∑
( )2 1
lim 11 42n
n nnh h
→∞
+ = + ×
Definite Integration
Mahesh Tutorials Science 35
( ) 2 2
11
lim 11 3 42n
nh n
→∞
+
= + ×
( )( )2 1 0
33 4 32
+ = + ×
( )33 2 9= +
= 33 + 18
= 51
2) ( )3
02x 3 dx+∫
Ans. Let, I = ( )3
02x 3 dx+∫
Let f(x) = 2x + 3, a = 0, b = 3
b a
nh
−= =
3 0
n
−
∴∴∴∴ nh = 3
∴∴∴∴ ( ) ( )3
01
f x dx lim f a + rhn
nr
h→∞
=
= ∑∫
( )1
lim f rhn
nr
h→∞
=
= ∑ [ ]a 0=∵
( )( )1
lim 2 rh 3n
nr
h→∞
=
= +∑
( )2
1
lim 2rh 3hn
nr
→∞=
= +∑
n n2
r 1 r 1
lim 2h r 3h 1n→∞
= =
= +
∑ ∑
( )2
1
1lim 2h 3h n
2
n
nr
n n
→∞=
+ = × + ×
∑
( )2 2 12h n 1n
lim 3nh2n→∞
× + = +
( ) ( )
( )2 2
2 3 1 03 3
2
× += +
= 9 + 9
= 18
3) ( )1
02x 3 dx+∫
Ans. Let, I = ( )1
02x 3 dx+∫
f(x) = 2x + 3, a = 0, b = 1
∴∴∴∴ ( )1
01
1f x dx lim f
n
nr
r
n n→∞=
=
∑∫
1
1 rlim 2 3
n
n
nr
n→∞=
= +
∑
1
1 2lim 3
n
nr
r
n n→∞=
= +
∑
21 1
2 3lim 1
n n
nr r
rnn→∞
= =
= +
∑ ∑
( )
2
n n 12 3lim n
2 nnn→∞
+ = × + ×
( )2
2
1n 12lim 3
2nn
n
→∞
+ = × +
( )2 1 0
32
+= +
= 1 + 3
= 4
4) ( )2
03x 5 dx+∫
Ans. Let,I = ( )2
03x 5 dx+∫
Let f(x) = 3x + 5, a = 0, b = 2
∴∴∴∴b a
hn
−= =
2 0
n
− = 2
n
∴∴∴∴ nh = 2
∴∴∴∴ ( ) ( )2
01
f x dx lim hf a rhn
nr
→∞=
= +∑∫
( )1
lim hf rhn
nr
→∞=
= ∑ [ ]a 0=∵
( )( )1
lim h 3 rh 5n
nr
→∞=
= +∑
n2
1 r 1
lim 3h r 5h 1n
nr
→∞= =
= +
∑ ∑
( )2 n n 1
lim 3h 5hn2n→∞
+ = × +
Definite Integration
36 Mahesh Tutorials Science
( )2 2 13h n 1lim 5hn
2n
n
→∞
× + = +
( ) ( )
( )2
3 2 1 05 2
2
+= +
= 6 + 10
= 16
5) ( )2 2
02x 5 dx+∫
Ans. Let I = ( )2 2
02x 5 dx+∫
Let f(x) = 2x2 + 5, a = 0, b = 2
∴∴∴∴b a
hn
−= =
2 0
n
− = 2
n
∴∴∴∴ n h = 2
∴∴∴∴ ( ) ( )2
01
f x dx lim hf a rhn
nr
→∞=
= +∑∫
( )1
lim hf rhn
nr
→∞=
= ∑ [ ]a 0=∵
( )( )2
1
lim h 2 rh 5n
nr
→∞=
= +∑
( )3 2
1
lim 2h r 5hn
nr
→∞=
= +∑
n n3 2
1 r 1 r 1
lim 2h r 5h 1n
nr
→∞= = =
= +
∑ ∑ ∑
( ) ( )3 n n 1 2n 1
lim 2h 5hn6n→∞
+ + = × +
( ) ( )( )
3 3 1 12h n 1 2n
lim 5 26n
n
→∞
× + + = +
( ) ( ) ( )3
2 2 1 0 2 010
6
+ += +
3210
6= +
= 46
3
6) ( )2 2
03x 5 dx+∫
Ans. Let, I = ( )2 2
03x 5 dx+∫
let f(x) = 3x2 + 5, a = 0, b = 2
∴∴∴∴ h = b a
n
− = 2 0
n
− =2
n
∴∴∴∴ nh = 2
( ) ( )2
01
f x dx lim hf a rhn
nr
→∞=
= +∑∫
( )1
lim hf rhn
nr
→∞=
= ∑ [ ]a 0=∵
( )( )2
1
lim h 3 rh 5n
nr
→∞=
= +∑
( )2 3
1
lim 3r h 5hn
nr
→∞=
= +∑
n3 2
1 1 r=1
lim 3h 5h 1n n
nr r
r
→∞= =
= +
∑ ∑ ∑
( ) ( )3 n n 1 2n 1
lim 3h 5h n6n→∞
+ + = × + ×
( ) ( )3 3 1 13h n 1 2lim 5nh
6n
n n
→∞
× + + = +
( ) ( ) ( )
( )3
3 2 1 0 2 05 2
6
+ += +
= 8 + 10
= 18
