8.2 Integration by parts

Formula for Integration by parts

( ) '( ) ( ) ( ) ( ) '( )

f x g x dx f x g x g x f x dx

udv uv vdu

• The idea is to use the above formula to simplify an integration task.• One wants to find a representation for the function to be integrated in the form udv so that the function vdu is easier to integrate than the original function.

• The rule is proved using the Product Rule for differentiation.

Deriving the Formula

Start with the product rule:

d dv duuv u v

dx dx dx

d uv u dv v du

d uv v du u dv

u dv d uv v du

u dv d uv v du

u dv d uv v du

u dv uv v du This is the Integration by Parts formula.

u dv uv v du

u differentiates to zero (usually).

dv is easy to integrate.

Choose u in this order: LIPET

Logs, Inverse trig, Polynomial, Exponential, Trig

Choosing u and v

Example 1:

cos x x dxpolynomial factor u x

du dx

cos dv x dx

sinv x

u dv uv v du LIPET

sin cosx x x C

u v v du

sin sin x x x dx

Example 2:

ln x dxlogarithmic factor lnu x

1du dx

x

dv dx

v x

u dv uv v du LIPET

lnx x x C

1ln x x x dx

x

u v v du

This is still a product, so we need to use integration by parts again.

Example 3:2 xx e dx

u dv uv v du LIPET

2u x xdv e dx

2 du x dx xv e u v v du

2 2 x xx e e x dx 2 2 x xx e xe dx u x xdv e dx

du dx xv e 2 2x x xx e xe e dx

2 2 2x x xx e xe e C

Example 4:

cos xe x dxLIPET

xu e sin dv x dx xdu e dx cosv x

u v v du sin sinx xe x x e dx

sin cos cos x x xe x e x x e dx

xu e cos dv x dx xdu e dx sinv x

sin cos cos x x xe x e x e x dx This is the expression we started with!

uv v du

Example 4(cont.):

cos xe x dxLIPET

u v v du

cos xe x dx 2 cos sin cosx x xe x dx e x e x

sin coscos

2

x xx e x e x

e x dx C

sin sinx xe x x e dx xu e sin dv x dx

xdu e dx cosv x

xu e cos dv x dx xdu e dx sinv x

sin cos cos x x xe x e x e x dx

sin cos cos x x xe x e x x e dx

This is called “solving for the unknown integral.”

It works when both factors integrate and differentiate forever.

Integration by Parts for Definite Integrals

12

0

arcsin x dx

b b b

b

aa a a

udv uv vdu u b v b u a v a vdu

2

Choose arcsin . Then

, and .1

u x

dxdv dx v x du

x

Formula

Integration by Parts Formula and the Fundamental Theorem of Calculus imply the above Integration by Parts Formula for Definite Integrals. Here we must assume that the functions u and v and their derivatives are all continuous.

Example

1 12 21

2

0 20 0

arcsin arcsin1

xdxx dx x x

x

Integration by Parts for Definite Integrals

12

2

20

Compute by the substitution 1 , 2 .1

1 30 1 and

2 4

xdxt x dt xdx

x

x t x t

331 42 4

20 1 1

3 2 31

2 221

xdx dtt

tx

Example (cont’d)

1 1 12 2 21

212

0 2 20 0 0

arcsinarcsin arcsin

21 1

xdx xdxx dx x x

x x

By the computations on the previous slide we now have

12 1

2

0

arcsin 2 3 2 3arcsin

2 2 12 2x dx

Combining these results we get the answer