ByDr. Safa Ahmed El-Askary

Faculty of Allied Medical of Sciences

Lecture (7&8)Integration by Parts

1

Integration by parts

udv uv vdu

d dv duuv u vdx dx dx

Product Rule:

d dv duuv dx u dx v dxdx dx dx

uv udv vdu

Integration by parts

udv uv vdu Let dv be the most complicated part of theoriginal integrand that fits a basic integrationRule (including dx). Then u will be the remaining factors.

Let u be a portion of the integrand whose derivative is a function simpler than u. Thendv will be the remaining factors (including dx).

OR

Integration by parts

udv uv vdu xxe dx u = x dv= exdx

du = dx v = ex

x x xxe dx xe e dx x x xxe dx xe e C

Integration by parts

udv uv vdu 2 lnx xdx u = lnx dv= x2dx

du = 1/x dx v = x3 /3

3 3 3 22 1ln ln ln

3 3 3 3x x x xx xdx x dx x dx

x

3 32 ln ln

3 9x xx xdx x C

Integration by parts

udv uv vdu arcsin xdx u = arcsin x dv= dx

2arcsin arcsin

1

xxdx x x dxx

v = x2

1

1du dx

x

2arcsin arcsin 1xdx x x x C

Integration by partsudv uv vdu

2 sinx xdx u = x2 dv = sin x dx

2 2sin cos 2 cosx xdx x x xdx du = 2x dx v = -cos x

u = 2x dv = cos x dxdu = 2dx v = sin x

2 2sin cos 2 sin 2sinx xdx x x x x xdx 2 2sin cos 2 sin 2cosx xdx x x x x x C

8.2 Trigonometric Integrals Powers of Sine and Cosine

sin cosn mu udusin cos cos sinn nu udu u udu

2 2sin 1 cosu u 1. If n is odd, leave one sin u factor and use

for all other factors of sin.

2 2cos 1 sinu u 2. If m is odd, leave one cos u factor and use

for all other factors of cos.

2 21sin (1 cos 1cos (1 cos2 ) 2 )22

oru u uu

3. If neither power is odd, use power reducing formulas:

Powers of sin and cos

2 2sin ( ) cos ( )d

3sin (2 )d

2 3sin ( ) cos ( )d

Powers of sin and cos3 2 2sin (2 ) sin 2 sin 2 (1 cos 2 )sin 2d d d

2 3 2 2 2 2sin cos sin cos cos sin (1 sin )cosd d d

2 3(sin 2 cos 2 sin 2 1 1cos 2 cos2 6

) 2 Cd

2 3 2 3(sin sin )cos (sin cos sin cos )d d 3 41 1sin sin

3 4C

Powers of sin and cos2 2 1 1sin ( )cos ( ) (1 cos 2 ) (1 cos 2 )

2 2d d

21 1 1(1 cos 2 ) (1 (1 cos 4 )4 4 21 1 1 1(1 cos 4 ) ( 4 )4 2 4 2

d d

d cos d

1 1 sin 48 16

C

Tangents and secants

sec tann mu udu

2tan sec sec sec tann nu udu u u udu

Create an integral that is shown above.2 2 2 2tan sec 1 tan 1 secor

4 4sec tan d

8.3 Eliminating radicals by trig substitution.

2 2a u 2 2a u 2 2u aPythagorean identities:

2 2cos 1 sin

2 2tan sec 1

2 2sec 1 tan

2 2a u Let u = a sin θ

2 2 2 2 2 2sin 1 sin cosa u a a a a

Trig Substitutions2 2a u2 2a u 2 2u a

2 29

dx

x x Let x = a sin θ = 3 sin θ

dx = 3 cos θ dθ

22 2 2 2

3cos 3cos9sin 3cos9 9sin 9 9sin

dx d d

x x

22

1 1 1csc cot9 9 9sin

d d C

Ex:

29 x

x3

21 99

x Cx

24 1

dx

x Let u=2x, a=1 so 2x = tan θ

dx = ½ sec2 θ dθ2

2

1 sec 1 sec2 sec 24 1

dx d dx

1 ln sec tan2

C

Ex:

24 1x 2x

1

21 ln 4 1 22

x x C

8.4 Partial Fractions2 5 6

dxx x

21

3 25 6A B

x xx x

21 ( 2) ( 3)

3 25 6A x B x

x xx x

1 ( 2) ( 3)A x B x If x = 2: 1=-B so B = -1If x =3: 1=A

21 1

3 25 6dx dx

x xx x

3ln | 3 | ln | 2 | ln2

xx x C or Cx

Partial Fractions-Repeated linear factors

2

3 2 25 20 6 6 1 9

12 ( 1)x x dx dx

x xx x x x

2 25 20 6 ( 1) ( 1)x x A x Bx x Cx If x =0: 6= A

If x = 1: 31=6(4)+2B+9, B = - 1

2

3 2 25 20 6

12 1

x x A B Cx xx x x x

If x = -1: -9 = -C, so C = 9

16ln | | ln | 1| 9( 1)x x x C

6 9ln1 1

x Cx x

2

3 25 20 6

2x x dxx x x

Quadratic Factors3

22 4 8( 1)( 4)

x x dxx x x

3

2 22 4 8

1( 1)( 4) 4x x A B Cx D

x xx x x x

3 2 22 4 8 ( 1)( 4) ( 4) ( ) ( 1)x x A x x Bx x Cx D x x

If x = 0 then A = 2If x = 1 then B = -2If x = -1 2 = -C +DIf x = 2 8 = 2C+D

Solving the system of equations you findC = 2 and D = 4.

3

2 2 22 4 8 2 2 2 4

1( 1)( 4) 4 4x x xdx dx

x xx x x x x

22ln | | 2ln | 1| ln( 4) 2arctan2xx x x C

3

2 28 13( 2)x xdxx

Repeated quadratic Factors3

2 2 2 2 28 13( 2) 2 ( 2)x x Ax B Cx Dx x x

3 28 13 ( )( 2)x x Ax B x Cx D 3 3 28 13 2 2x x Ax Ax Bx B Cx D

3 2 3 200 13 2 28 Bx x x x x AxA Cx D B

A=8

13=2A+C

For third degree: For second degree: B=0

For first degree:

For constant: D+2B=0

3

2 28 13( 2)x xdxx

Repeated quadratic Factors

3

2 2 2 2 28 13 8 3( 2) 2 ( 2)x x x xdx dxx x x

32

2 2 28 13 34ln( 2)( 2) 2( 2)x xdx x Cx x

A=8

13=2A+C

B=0

D+2B=0So, D=0 and C = -3

8.8 Areas under curves with infinite domain or range

21

ln xdxx

1

0

1 dxx

Improper Integrals with infinite limits

Upper limit infinite

Lower limit infinite

Both limit infinite

2 21 1

ln lnlimb

bx xdx

x x

Infinite limits

2 21 1

ln lnlimb

bx xdx dx

x x

2

ln

1

dxu x dux

dxdv vxx

2 2ln ln ln 1x x dx xdx

x x xx x

We say the improper integral CONVERGES toThe value of 1. (The area is finite.)

Evaluation

Use L’Hôpital’s rule

211

ln ln 1lim limb b

b bx xdx

x xx

ln 1lim (0 1) 0 0 0 1bb

b b

When both limits are infinite

20

2

0

2

2 00

2 2

2

0 00

0

1

lim lim arctan 02

1

lim lim arctan 02 21

2

1

2

2

1

1 1

a a a

bb

ab

b

dxx

dxx

dx dx xx

dx dx xx x

x

x

dx

Improper Integrals-integrand becomes infinite

upper endpoint

lowerendpoint

interior point

1 1

00

10

0

1 1lim

lim 2

lim 2 2 2

aa

a a

a

dx dxx x

x

a

Integrals with Infinite discontinuities.

The integral converges to 2.

1

20 3

1 131 12 2 0

0 03 3

131

3

32 2

3

0 3 11

lim lim 3 11 1

lim 3( )

1

1 3 3

1b

bb b

b

dx

x

dx dx xx x

b

d d

x x

x x

33 3 131 12 2

1 3 3

133

1

3

lim lim 3 11 1

1lim 3(3 1) 3(

3

1) 3 23

3 2The total integral is

c cc c

c

dx dx xx x

c

Calculation with infinite discontinuity

Area is finite

Integral converges to 1

21

1t

dxx

Area is infinite

Integral diverges

1

1t

dxx

21

) dxx

a

1

)b dxx

11 2

)c dx

x

Integrals of the form 1

pdxx

211

1 1) lim lim lim ( 1) 1b b

b b bdxa

x bx

11

) lim lim ln lim ln ln1b

bb b b

dxb x bx

1 12 2

11 2 1

) lim lim lim (1)

bb

b b bdxc x b

x

Convergence or divergence

Integrals of the form 1

pdxx

Converge if p > 1 and diverge if p = 1 or p < 1.

31 2

dx

x

21 3

dx

x

1

3dxx

51

4dxx

Which of the following converge and which diverge?

Direct comparison testIf f and g are continuous functions with f(x) g(x)For all x a. Then…..

( )a

f x dx

Converges if ( )a

g x dx

Converges

( )a

g x dx

Diverges if ( )a

f x dx

Diverges

A function converges if its values are smaller than another function known to converge.

A function diverges if its values are larger than another function known to diverge.

Limit Comparison test for convergenceIf f and g are positive and continuous on [a, )And if ( )lim , 0

( )xf x L Lg x

Then the integrals ( )a

f x dx

( )a

g x dx

and

If

If

both converge or both diverge:

( )a

g x dx

diverges and

then

( )lim( )x

f xg x

( )a

f x dx

also diverges.