Integration by Parts

I. Basic Technique

By the Product Rule for Derivatives, { } )()()()()()( xfxgxgxfxgxfdxd

ʹ′+ʹ′= . Thus,

[ ] ∫ ∫∫ =ʹ′+ʹ′⇒=ʹ′+ʹ′ dxxfxgdxxgxfxgxfdxxfxgxgxf )()()()()()()()()()(

∫ ∫ ʹ′−=ʹ′⇒ dxxfxgxgxfdxxgxfxgxf )()()()()()()()( . This formula for integration

by parts often makes it possible to reduce a complicated integral involving a product to a simpler integral. By letting dxxfduxfu )()( ʹ′=⇒= )()( xgvdxxgdv =⇒ʹ′= we get the more common formula for integration by parts:

∫ ∫−= vduuvudv

Example 1: Find ∫ xdxx ln .

Let xu ln= and dxx

duxdxdv 1=⇒= and 2

21 xdxxv ∫ == . Thus,

−⎟⎠

⎞⎜⎝

⎛=−=== ∫ ∫ ∫∫ 2

21)(ln))((lnln xxvduuvudvdxxxxdxx

=+⎟⎠

⎞⎜⎝

⎛−=−=⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ ∫∫ Cxxxdxxxxdxx

x 2222

21

21ln

21

21ln

211

21

Cxxx +− 22

41ln

21 .

It is possible that when you set up an integral using integration by parts, the resulting integral will be more complicated than the original integral. In this case, change your substitutions for u and dv.

1

Example 2: Find ∫ dxxx sin .

Let xu sin= and dxxdudxxdv cos=⇒= and 2

21 xdxxv ∫ == . Thus,

−⎟⎠

⎞⎜⎝

⎛=−=== ∫ ∫ ∫∫ 2

21)(sin))((sinsin xxvduuvudvdxxxdxxx

dxxxdxxx cos21)(cos

21 22 ∫∫ =⎟

⎠

⎞⎜⎝

⎛ . Notice that this resulting integral is

more complicated that the original one. Therefore, let xu = and =dv

dxdudxx =⇒sin and ∫ −== xdxxv cossin . Thus, ∫ dxxx sin =

∫ ∫ ∫ ∫ =+−=−−−=−= xdxxxdxxxxvduuvudv coscos)cos()cos)((

Cxxx ++− sincos .

Example 3: Find ∫ dxxln .

Let xu ln= and dxx

dudxdv 11 =⇒= and ∫ == xdxv 1 . Thus,

=⎟⎠

⎞⎜⎝

⎛−=−=== ∫ ∫ ∫ ∫∫ )(1))((ln)1)((lnln xdxx

xxvduuvudvdxxdxx

∫ +−=− Cxxxdxxx ln1ln .

Example 4: Find ∫ dxxarctan .

As in the previous example, let xu arctan= and dxx

dudxdv 211+

=⇒=

and ∫ == xdxv 1 . Thus, ∫ ∫∫ =−== vduuvudvdxxarctan −xxarctan

=+

−=+

−=⎟⎠

⎞⎜⎝

⎛+ ∫∫∫ dx

xxxxdx

xxxxdx

xx 222 1

221arctan

1arctan

11

Cxxx ++− 1ln21arctan 2 = Cxxx ++− )1ln(

21arctan 2 .

2

Example 5: Find dxxx

∫ 2

ln .

Let xu ln= and dxx

dudxxdxx

dv 11 22 =⇒== − and =−== −−∫ 12 1xdxxv

x1− . Thus, −⎟

⎠

⎞⎜⎝

⎛ −=−==⎟⎠

⎞⎜⎝

⎛= ∫ ∫∫∫ xxvduuvudvdx

xxdx

xx 1)(ln1)(lnln

22

=+−−

=+−

=+−

=⎟⎠

⎞⎜⎝

⎛ −⎟⎠

⎞⎜⎝

⎛ ∫∫∫ − Cxx

xdxxxxdx

xxx

xdxx

1lnln1ln11 22

Cxx

+−− 1ln .

Sometimes, it is necessary to use integration by parts more than once.

Example 6: Find dxxx cos2∫ .

Let 2xu = and xdxdudxxdv 2cos =⇒= and ∫ == xdxxv sincos . Thus,

∫∫ ∫∫ −=−=−== xxxxdxxxvduuvudvdxxx sin))(sin2())(sin(cos 222

∫ xdxx sin2 . Notice that the resulting integral ∫ dxxx sin is less complicated

than the original one, but integration by parts is needed to evaluate it.

Let xu = and dxdudxxdv =⇒= sin and ∫ −== xdxxv cossin . Thus,

∫ ∫ ∫∫ +−=−−−=−== xxdxxxxvduuvudvdxxx cos)cos()cos(sin

∫ +−= xxxdxx sincoscos . Finally, we get −=∫ xxdxxx sincos 22

[ ] CxxxxxCxxx +−+=++− sin2cos2sinsincos2 2 . [Note: We will be

able to integrate dxxx cos2∫ more easily with tabular integration; this

technique will be described later.] The next illustration of repeated integration by parts deserves special attention.

Example 7: Find dxxe x sin∫ .

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Let xeu = and dxedudxxdv x=⇒= sin and xxdxv cossin −== ∫ . Thus,

∫∫ ∫∫ =−−−=−== ))(cos()cos)((sin dxexxevduuvudvdxxe xxx

xdxexe xx coscos ∫+− . Notice that integration by parts is now needed to

evaluate xdxe x cos∫ . Let xeu = and dxeduxdxdv x=⇒= cos and

∫ == xxdxv sincos . Thus, −=−== ∫ ∫∫ xevduuvudvxdxe xx sincos

xdxexedxex xxx sinsin))((sin ∫∫ −= . Returning to the original problem,

dxxe x sin∫ = ∫∫ −+−=+− dxxexexexdxexe xxxxx sinsincoscoscos .

Thus, ∫ ⇒+−= xexedxxe xxx sincossin2 dxxe x sin∫ = +− xex cos21

Cxex +sin21 and this does check.

The next example illustrates an interesting type of integral that surprisingly requires integration by parts.

Example 8: ( ) dxx∫ sin .

Let ⇒=⇒=⇒= dxuduxuxu 22 ( ) ∫∫ == )2)((sinsin uduudxx

∫ duuu sin2 . In example 2, we got the following using integration by parts:

∫ dxxx sin = Cxxx ++− sincos or ∫ duuu sin = Cuuu ++− sincos .

Thus, ( ) dxx∫ sin = ∫ duuu sin2 = ( )+−=++− xxCuuu cos2sin2cos2

( ) Cx +sin2 and this does check.

In general, to evaluate ( ) dxxf n∫ , let ⇒=⇒=⇒= − dxdunuxuxu nnn 1

( ) dxxf n∫ = ∫ − duufun n )(1 .

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II. Tabular Integration

Integrals of the form ∫ dxxgxf )()( , in which f can be differentiated repeatedly

to become zero and g can be integrated repeatedly without difficulty, can be evaluated using tabular integration.

Example: Find dxex x23∫ .

)(xf and its derivatives )(xg and its antiderivatives 3x (+) xe 2

23x (–) xe221

6x (+) xe241

6 (–) xe281

0 xe2161

dxex x23∫ = Cexeexex xxxx +−+−+ 222223

166

86

43

21

= Cexeexex xxxx +−+− 222223

83

43

43

21

III. Reduction Formulas Integration by parts can be used to derive reduction formulas for integrals. These are formulas that express an integral involving a power of a function in terms of an integral that involves a lower power of that function

5

Example 1: Prove the reduction formula dxxnxxdxx nnn ∫∫ −−= 1)(ln)(ln)(ln and

use the result to find dxx∫ 2)(ln .

Let nxu )(ln= and ⎟⎠

⎞⎜⎝

⎛=⇒= − dxx

xndudxdv n 1)(ln 1 and ∫ == xdxv 1 . Thus,

−=⎟⎠

⎞⎜⎝

⎛−=−== ∫∫ ∫ ∫ − nnnn xxdxx

xnxxxvduuvudvdxx )(ln1)(ln)()(ln)(ln 1

dxxn n∫ −1)(ln . Thus, −−=−= ∫∫ )(ln[2)(lnln2)(ln)(ln 222 xxxxdxxxxdxx

Cxxxxxdx ++−=∫ 2ln2)(ln]1 2 and this checks.

Example 2: Prove the reduction formula +−= −∫ xxn

dxx nn cos)(sin1)(sin 1

dxxnn n∫ −− 2)(sin1 .

Let 1)(sin −= nxu and dxxxndudxxdv n cos))(sin1(sin 2−−=⇒= and v =

∫ −= xdxx cossin . Thus, ∫∫∫ === − udvdxxxdxx nn )(sin)(sin)(sin 1

=−−−−=− ∫∫ −− dxxxnxxxvduuv nn cos))(sin1)(cos()cos()(sin 21

+−=−+− −−− ∫ xxdxxxnxx nnn cos)(sincos)(sin)1(cos)(sin 1221

∫∫ −−+−=−− −−− dxxnxxdxxxn nnn 2122 )(sin)1(cos)(sin)sin1()(sin)1(

∫∫∫ ⇒−+−=⇒− −− dxxnxxdxxndxxn nnnn 21 )(sin)1(cos)(sin)(sin)(sin)1(

∫∫ −− −+−= dxxnnxx

ndxx nnn 21 )(sin1cos)(sin1)(sin .

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Practice Sheet for Integration by Parts

(1) xx ln3∫ dx =

(2) Prove the following reduction formula: −+

= +∫ nmnm xxm

dxxx )(ln1

1)(ln 1

dxxxmn nm 1)(ln1

−∫+ for 1−≠m .

(3) ∫ xarcsin dx =

(4) ( )∫ xcos dx =

(5) ∫ 3lnxx dx =

(6) ∫ x arcsec x dx =

(7) xex cos∫ dx =

(8) ( )23 sin xx∫ dx =

(9) ∫ xe dx =

(10) xex 32∫ dx =

(11) xx sin2∫ dx =

7

(12) =∫ dxe x3

(13) =∫ dxx

e x

3

3

(14) ∫ =dxx)sin(ln

Solution Key for Integrations by Parts

(1) Let xu ln= and ⇒= dxxdv 31

dxx

du 1= and ⇒== ∫ 3

431

43 xdxxv

xx ln3∫ dx = −=−=⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧− ∫∫ 3

431

34

34

34

43

43

431

43ln

43 xdxxxdx

xxxx

Cx +34

169 .

(2) Let nxu )(ln= and ⎟⎠

⎞⎜⎝

⎛=⇒= − dxx

xndudxxdv nm 1)(ln 1 and ∫ == dxxv m

⇒+

+1

11 mxm

( ) =⎟⎠

⎞⎜⎝

⎛+

⎟⎠

⎞⎜⎝

⎛−+

= +−+ ∫∫ 111

111)(ln)(ln1ln mnnmnm xm

dxx

xnxxxm

dxxx

∫ −+

+−

+dxxx

mnxx

xmnmnm 11 )(ln

1)(ln1 .

(3) Let xu arcsin= and dxx

dudxdv21

1−

=⇒= and ∫ ⇒== xdxv 1

∫ xarcsin dx = =−

−+=

−− ∫∫ dx

xxxxdx

xxxx

22 12

21arcsin

1arcsin

.1arcsin 2 Cxxx +−+

8

(4) Let ⇒=⇒=⇒= dxwdwxwxw 22 ( )∫ xcos dx = ∫ dwwwcos2 . Let

wu = and dwdudwwdv =⇒= cos and ∫ ⇒== wdwwv sincos ( )∫ xcos dx =

∫ dwwwcos2 = ∫ ∫ =−=− dwwwwdwwww sin2sin2)(sin2)(sin2

CxxxCwww ++=++ )cos(2)sin(2cos2sin2 .

(5) Let xu ln= and dxx

dudxx

dv 113 =⇒= and ⇒−== −−∫ 23

21 xdxxv

∫ 3lnxx dx = =+−=⎟

⎠

⎞⎜⎝

⎛+− ∫∫ −−−− dxxxxdxxx

xx 3222

21ln

211

21ln

21

Cxx

xCxxx +−−

=+−− −−22

22

41

2ln

41ln

21 .

(6) Let xarcu sec= and dxxx

dudxxdv1

12 −

=⇒= and ⇒== ∫ 2

21 xdxxv

∫ x arcsec x dx = Cxxarcxdxxxxarcx +−−=−

− ∫ 121sec

21

121sec

21 22

2

2 .

(7) Let xu cos= and dxxdudxedv x sin−=⇒= and ⇒== ∫ xx edxev =∫ dxxex cos

xdxexe xx sincos ∫+ . Use integration by parts to evaluate ∫ dxxe x sin . Let

xu sin= and dxxdudxedv x cos=⇒= and ⇒== ∫ xx edxev ∫ dxxe x sin =

dxxexe xx cossin ∫− . Thus, xex cos∫ dx = xdxexe xx sincos ∫+ =

⇒−+=⎭⎬⎫

⎩⎨⎧

−+ ∫∫ dxxexexedxxexexe xxxxxx cossincoscossincos

⇒+=∫ xexedxxe xxx sincoscos2 ( ) Cxxedxxex

x ++=∫ sincos2

cos .

9

(8) Let ( ) ( ) ==⇒=⇒= ∫∫ dxxxxdxxxdxxdwxw 2sin21sin2 22232

∫ dwwwsin21 . Let wu = and dwdudwwdv =⇒= sin and ∫ == dwwv sin

+−=⎭⎬⎫

⎩⎨⎧

+−=⇒− ∫∫ wwdwwwwdwwww cos21coscos

21sin

21cos

( ) ( ) CxxxCw ++−=+ 222 sin21cos

21sin

21 .

(9) Let ⇒=⇒=⇒= dxdwwxwxw 22 ∫ xe dx = dwwew∫2 . Let wu = and

dwdudwedv w =⇒= and ⇒== ∫ ww edwev =−= ∫∫ dwewedwwe www

⇒− ww ewe ∫ xe dx = CeexCewedwwe xxwww +−=+−=∫ 22222 .

(10) Use tabular integration: f (x) and its derivatives g (x) and its antiderivatives

2x xe3 (+)

x2 xe331

(–)

2 xe391

(+) 0

xe3271

Thus, xex 32∫ dx = Cexeex xxx ++− 3332

272

92

31 .

10

(11) Use tabular integration: f (x) and its derivatives g (x) and its antiderivatives

2x xsin (+) x2 xcos− (–) 2 xsin− (+) 0 xcos

Thus, xx sin2∫ dx = Cxxxxx +++− cos2sin2cos2 .

(12) Let ⇒=⇒=⇒= dxdwwxwxw 233 3 =∫ dxe x3 dwew w∫ 23 . dwew w∫ 2 can

integrated using tabular integration: f (x) and its derivatives g (x) and its antiderivatives 2w (+) we w2 (–) we 2 (+) we 0 we

dwew w∫ 2 = ⇒++− Ceweew www 222 == ∫∫ dwewdxe wx 233

333

663 33 2 xxx eexex +− + C.

(13) Let ⇒=⇒=⇒= dxdwwxwxw 233 3 =∫ dxx

e x

3

3

∫∫ = dwwedwwew ww

332

.

In solution to problem #9 previously, ⇒−=∫ www ewedwwe =∫ dxx

e x

3

3

11

CeexCewedwwe xxwww +−=+−=∫33

33333 3 .

(14) Let =u sin(ln x) and ⎟⎠

⎞⎜⎝

⎛=⇒= dxx

xdudxdv 1)cos(ln and ∫ ⇒== xdxv 1

∫ =dxx)sin(ln ∫− dxxxx )cos(ln)sin(ln . Use integration by parts to evaluate

∫ dxx)cos(ln . Let )cos(ln xu = and ⎟⎠

⎞⎜⎝

⎛−=⇒= dxx

xdudxdv 1)sin(ln and

∫ ⇒== xdxv 1 ∫ dxx)cos(ln = ∫+ dxxxx )sin(ln)cos(ln . Thus, ∫ =dxx)sin(ln

∫− dxxxx )cos(ln)sin(ln = −=⎭⎬⎫

⎩⎨⎧

+− ∫ )sin(ln)sin(ln)cos(ln)sin(ln xxdxxxxxx

∫ ∫ ⇒−=⇒− )cos(ln)sin(ln)sin(ln2)sin(ln)cos(ln xxxxdxxdxxxx

∫ =dxx)sin(ln Cxxxx +− )cos(ln21)sin(ln

21 .

12