Integration by PartsIntegration by Parts

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

This is the Integration by Parts formula.

dxvuuvdxvu

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

u differentiates to zero (usually).

v’ is easy to integrate.

Choose u in this order: LIPET

Logs, Inverse trig, Polynomial, Exponential, Trig

dxvuuvdxvu

Example 1:

polynomial factor u x

sinv x

LIPET

sin sin x x x dx

dxxx cos

xv cos

dxvuuvdxvu

1u dxvuuvdxvu

cxxxdxxx cossincos

Example 2:

logarithmic factor

LIPET

dxxln

1v

dxvuuvdxvu

xu

1

dxvuuvdxvu

cxxxdxx ln1ln

xu ln

xv

1ln x x x dx

x

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

Example 3:2 xx e dx

LIPET

2u xxv e

u xxv e

dxvuuvdxvu

xu 2

xev

1u

xev

dxvuuv

dxxeex xx 22

dxxeex xx 22

dxexeex xxx 22

cexeexdxex xxxx 2222

dxvuuvdxvu

u dv uv v du

The Integration by Parts formula can be written as:

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 continued …

cos xe x dx LIPET

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 4 continued …

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:

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

Example 3:2 xx e dx

LIPET

2u xxv e

u xxv e

dxvuuvdxvu

xu 2

xev

1u

xev

dxvuuv

dxxeex xx 22

dxxeex xx 22

dxexeex xxx 22

cexeexdxex xxxx 2222

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

Acknowledgement

I wish to thank Greg Kelly from Hanford High School, Richland, USA for his hard work in creating this PowerPoint.

http://online.math.uh.edu/

Greg has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.

Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au