6.3 Integration By Parts

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

The Integration by Parts formula is a “product rule” for integration.

u differentiates to zero (usually).

dv is easy to integrate.

Choose u in this order: LIPET

Logs, Inverse trig, Polynomial, Exponential, Trig

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:

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 4:2 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 5:

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 6:

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 xe 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

Example 6:

cos xe x dx u v v du

This is called “solving for the unknown integral.”

It works when both factors integrate and differentiate forever.

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

sin coscos

2

x xx e x e xe x dx C

sin sinx xe x x e dx

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

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

A Shortcut: Tabular Integration

Tabular integration works for integrals of the form:

f x g x dx

where: Differentiates to zero in several steps.

Integrates repeatedly.

2 xx e dx & deriv.f x & integralsg x

2x

2x

2

0

xexexexe

2 xx e dx 2 xx e 2 xxe 2 xe C

Compare this with the same problem done the other way:

Example 5:2 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 This is easier and quicker to do with tabular integration!

3 sin x x dx3x

23x

6x

6

sin x

cos x

sin xcos x

0

sin x

3 cosx x 2 3 sinx x 6 cosx x 6sin x + C

p