Integration by Parts - MR. SOLIS'...
Transcript of Integration by Parts - MR. SOLIS'...
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.
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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
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Cxxx ++− 1ln21arctan 2 = Cxxx ++− )1ln(
21arctan 2 .
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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
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43
43
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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
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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 =
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(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 +−+
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(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 .
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(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 .
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(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 .
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