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Integration by Parts

dx

dvu

dx

duv

dx

dy

If ,uvy where u and v are

both functions of

x. Substituting for

y,

xxxdx

xxdcos1sin

)sin(

e.g. If , xxy sin

We develop the formula by considering how to differentiate products.

dx

dvu

dx

duv

dx

uvd

)(


dxxxdxxxx cossinsin

xxxdx

xxdcos1sin

)sin(So,

Integrating this equation, we get

dxxxdxxdxdx

xxd cossin)sin(

The l.h.s. is just the integral of a derivative, so, since integration is the reverse of differentiation, we get

Can you see what this has to do with integrating a product?


dxxxdxxxx cossinsin

Here’s the product . . .

if we rearrange, we get

dxxxxdxxx sinsincos

The function in the integral on the l.h.s. . . .. . . is a product, but the one on the r.h.s. . .

. is a simple function that we can integrate easily.


dxxxdxxxx cossinsin

Here’s the product . . .

if we rearrange, we get

dxxxxdxxx sinsincos

Cxxx )cos(sin

Cxxx cossin

We need to turn this method ( called integration by parts ) into a formula.

So, we’ve integrated !

xx cos


Integrating:

dxdx

dvudx

dx

duvdx

dx

uvd )(

dxdx

dvudx

dx

duvuv

dxdx

duvuvdx

dx

dvu

dx

dvu

dx

duv

dx

uvd

)(xxx

dx

xxdcossin

)sin(

dxxxdxxdxdx

xxd cossin)sin(

Rearranging:

Simplifying the l.h.s.:

dxxxdxxxx cossinsin

dxxxxdxxx sinsincos

Generalisation

Example


SUMMARY

dxdx

duvuvdx

dx

dvu

To integrate some products we can use the formula



dx

duto get . . .

dxdx

duvuvdx

dx

dvu

Using this formula means that we differentiate

one factor, u

So,


So,

dxdx

duvuvdx

dx

dvu

and integrate the other , dx

dvto get v


one factor, u dx

duto get . . .


So,

dxdx

duvuvdx

dx

dvu

e.g. 1 Find dxxx sin2

xu 2 xdx

dvsinand

2dx

duxv cos

differentiate

integrate

and integrate the other , dx

dvto get v


one factor, u dx

duto get . . .

Having substituted in the formula, notice

that the 1st term, uv, is completed but the

2nd term still needs to be integrated.

( +C comes

later )


)cos(2 xx dxx 2)cos(

We can now substitute into the formula

So,xu 2

2dx

du xv cosdifferentiate

integrate

xdx

dvsinand

udx

dv u v vdx

du

dxdx

duvuvdx

dx

dvu

dxxx sin2


C

So,

dxdx

duvuvdx

dx

dvu

)cos(2 xx dxx 2)cos(

xsin2

xu 2

2dx

du xv cosdifferentiate

integrate

We can now substitute into the formula

xdx

dvsinand

dxxcos2xx cos2The 2nd term needs integrating

xx cos2

dxxx sin2


dxdx

duvuvdx

dx

dvu

2

)(2 xe

x

dx

e x

12

2

xu

1dx

dudifferentiate

integrate

xedx

dv 2and

dxexe xx

22

22

e.g. 2 Find dxxe x2

Solution:

v2

2 xe

Cexe xx

42

22

dxxe x2So,

This is a compound function, so we must be careful.


ExercisesFind dxxe x

1. dxxx 3cos2.

dxexe xx

Cexe xx

dxxx 3cos2. dxxx

x

3

3sin

3

3sin)(

Cxxx

9

3cos

3

3sin

1.

Solutions: dxxex

Integration by PartsDefinite Integration by PartsWith a definite integral it’s often easier to do the indefinite integral and insert the limits at the end.

1

0dxxe x

dxxe x dxexe xx

1

0xx exe

Cexe xx

0011 01 eeee

10 1

We’ll use the question in the exercise you have just done to illustrate.


Integration by parts cannot be used for every product.

Using Integration by Parts

It works if we can integrate one factor of the

product, the integral on the r.h.s. is easier* than

the one we started with.

* There is an exception but you need to learn the general rule.


Solution:

What’s a possible problem?

Can you see what to do?

If we let and , we will need to

differentiate and integrate x.

xu ln xdx

dv

xln

ANS: We can’t integrate .

xln

xlnTip: Whenever appears in an integration by parts we choose to let it

equal u.

dxdx

duvuvdx

dx

dvue.g. 3 Find dxxx ln


dxdx

duvuvdx

dx

dvu

So, xu ln xdx

dv

xdx

du 1

2

2xv

dxxx ln xx

ln2

2

dxx

x 1

2

2

The r.h.s. integral still seems to be a product!BUT . . .

xx

dxxx ln2

ln2

4

2xC

x

cancels.

e.g. 3 Find dxxx ln

differentiate

integrate

So, xx

dxxx ln2

ln2

dxx

2


e.g. 4 dxex x 2

Solution:

dxdx

duvuvdx

dx

dvu

Let 2xu xedx

dv

xdx

du2 xev

dxxeexdxex xxx 222

dxxeexdxex xxx 222

22

1 IexI x

The integral on the r.h.s. is still a product but using the method again will give us a simple function.We write

and

1I 2I


e.g. 4 dxex x 2

Solution:

2dx

duxev

22

1 IexI x

2ISo, xxe2 dxe x2

xxe2 dxe x2

Cexe xx 22Substitute in

( 1 )

. . . . . ( 1 )

xx exdxex 22

Let xu 2 xedx

dv and dxxeI x22

Cexe xx 22


dxdx

duvuvdx

dx

dvu

Solution:It doesn’t look as though integration by parts will help since neither function in the product gets easier when we differentiate it.

e.g. 5 Find dxxe x sin

However, there’s something special about the 2 functions that means the method does work.

Example


dxdx

duvuvdx

dx

dvu

xeu xdx

dvsin

xedx

du xv cos


dxxe x sin xe x cos dxxe x cos

xe x cos dxxe x cos

Solution:

We write this as: 21 cos IxeI x


dxdx

duvuvdx

dx

dvu

xeu xdx

dvcos

xedx

du xv sin


xe x sin dxxe x sin

dxxeI x sin1where

So, 21 cos IxeI x

and dxxeI x cos2

We next use integration by parts for I2

dxxe x cos

12 sin IxeI x


dxdx

duvuvdx

dx

dvu

xeu xdx

dvcos

xedx

du xv sin


dxxeI x sin1

xe x sin dxxe x sin

So,where

21 cos IxeI x

and dxxeI x cos2

We next use integration by parts for I2

dxxe x cos

12 sin IxeI x


dxdx

duvuvdx

dx

dvue.g. 5 Find dxxe x sin

So, 21 cos IxeI x

2 equations, 2 unknowns ( I1 and I2 )

!Substituting for I2 in

( 1 )

. . . . . ( 1 )

. . . . . ( 2 )

xeI x cos1

2I 1sin Ixe x


dxdx

duvuvdx

dx


So, 21 cos IxeI x



( 1 )

. . . . . ( 1 )

. . . . . ( 2 )

xeI x cos1 1sin Ixe x

2I 1sin Ixe x


dxdx

duvuvdx

dx


So, 21 cos IxeI x



( 1 )

. . . . . ( 1 )

. . . . . ( 2 )

xeI x cos1

xexeI xx sincos2 1

2

sincos1

xexeI

xx C

1sin Ixe x

2I 1sin Ixe x

Integration by PartsExercises

2.dxxx sin2( Hint: Although 2. is not a product it can be

turned into one by writing the function as . )

xln1

1. dxx2

1ln

Integration by PartsSolutions:

dxxx sin21.

xdx

du2 xv cos

2xu xdx

dvsinan

dLet

1I

dxxxxxI cos2cos21

dxxxxxI cos2cos21

2I

2dx

duxv sin

dxxxxI sin2sin22

Cxxx cos2sin2

Cxxxxxdxxx cos2sin2cossin 22

xu 2 xdx

dvcosan

dLetFor I2:

. . . . . ( 1 )

Subs. in ( 1 )


2.

xdx

du 1 xv

xu ln 1dx

dvand

Let

dxxln1 xx ln dxx

x 1

xx ln dx1

Cxxx ln

This is an important application of integration by parts

dxx2

1ln dxx

2

1ln1

dxx2

1lnSo, 2

1ln xxx

11ln122ln2 12ln2


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