the method of integration by parts. main idea if u & v are differentiable functions of x, then...
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The Method of Integration by Parts
Main Idea
If u & v are differentiable functions of x, then
By integrating with respect to x, we get:
')'('
'')'(
vuuvuv
vuuvuv
vduuvudv
dxvudxuvdxuv ')'('
When to use this method?
When the integrand is a product of the form udv, such that we do not know how to find the integral ∫udv, but can find v = ∫dv and the integral ∫vdu.
Examples I
When, we have an integrand, similar to one of the following: ( where b and c are any real numbers)
1. xn cos(cx) or xn sin(cx) ; where n is a natural number
2. xn ecx or xn acx
; where n is a natural number and b is a base for an exponential function ( b is positive and not equal to 1)
3. x lnx or xc lnxb
Example 1
dxxxI cos
cxxx
dxxxx
vduuvI
xvdxdu
dxxdvxuLet
dxxxI
cossin
sinsin
sin,..
cos,
cos
Example 2
dxxxI sin2
cxxxxx
cxxxxx
dxxxxx
vduuvI
xvxdxdu
dxxdvxuLet
dxxxI
cos2sin2cos
]cossin[2cos
cos2cos
cos,2
sin,
sin
2
2
2
2
2
Example 3
dxxxI cos3
cxxxxxxx
dxxxxx
vduuvI
xvdxxdu
dxxdvxuLet
dxxxI
]cos2sin2cos[3sin
sin3sin
sin,3
cos,
cos
23
23
2
3
3
Example 5
dxxeI x
cexe
dxexe
vduuvI
evanddxdu
dxedvandxuLet
dxxeI
xx
xx
x
x
x
Example 5
dxexI x2
cexeex
cexeex
dxxeex
vduuvI
evandxdxdu
dxedvandxuLet
dxexI
xxx
xxx
xx
x
x
x
22
][2
2
2
2
2
2
2
2
Example 6
dxexI x3
cexeexex
cexeexex
dxexex
vduuvI
evanddxxdu
dxedvandxuLet
dxexI
xxxx
xxxx
xx
x
x
x
663
]22[3
3
3
23
23
23
2
3
3
Example 7
dxxxI ln
cxx
cx
cxdxx
dxx
vduuvI
vanddxdu
xdxdvandxuLet
dxxxI
x
xx
x
xxx
xx
241
2
221
2
21
2
122
21
ln
ln
ln
ln
ln
ln
2
22
2
22
2
Example 8
dxIxx)ln(
5 35 3
11
153
53
531
1
11
ln
,
ln)ln()ln(
,
)ln(
53
53
5 3
53
5 3
5 35 3
IdxxxI
Therfore
xx
x
simplifyFirst
dxI
x
x
xx
cxxx
cxx
dxxxx
dxxxx
vduuvI
xvanddxdu
dxxdvandxuLet
dxxxI
x
x
xx
52
52
52
52
52
53
52
52
52
52
52
52
53
53
425
25
25
25
25
25
125
25
1
251
1
ln
ln
ln
ln
ln
ln
cxxx
cxxxI
Therefore
52
52
52
52
415
23
425
25
53
ln
]ln[
,
Example 9
dxxI ln
cxxx
cdxxx
dxx
xxx
vduuvI
xvanddxx
du
dxdvandxuLet
dxxI
ln
ln
1ln
1
ln
ln
Examples II:Integrals valued by
Repeated Use of the Method When, we have an integrand, similar to one
of the following: ( where b and c are any real numbers)
1. sin(bx) cos(cx)
2. ecx sin(bx) or ecx cos(bx)
Example 1
dxxeI x cos
)2(..............................22sin
2cos22sin
2cos2
2sin:
2sin
,
)1.......(....................22cos
2sin22cos
2sin2
2cos
2cos
1
1
1
Ixe
dxexxeI
evanddxxdu
dxedvandxuLet
dxexI
Where
Ixe
dxexxe
vduuvI
evanddxxu
dxedvandxuLet
dxxeI
x
xx
x
x
x
x
xx
x
x
x
]2sin22cos[
2sin22cos5
42sin22cos
]22sin[22cos
)1()2(
)2.......(....................22sin
&
)1.......(....................22cos
,
51
1
1
1
xexeI
xexeI
IxexeI
IxexeI
infromISubstitute
IxeI
IxeI
haveWe
xx
xx
xx
xx
x
x
Example 2
dxxxI cos2cos
)2(..............................2cos2sin
cos2cos2cos2sin
cos.2cos2..
sin2sin:
sin2sin
,
)1.......(....................2sin2cos
sin2sin2sin2cos
sin2sin2
cos2cos
cos2cos
1
1
1
Ixx
dxxxxxI
xvanddxxdu
dxxdvandxuLet
dxxxI
Where
Ixx
dxxxxx
vduuvI
xvanddxxdu
dxxdvandxuLet
dxxxI
.
sin2cos3
1cos2sin
3
2
,
sin2cos3
1cos2sin
3
2
]sin2coscos2sin2[3
1
sin2coscos2sin23
4cos2sin2sin2cos
]2cos2sin[2sin2cos
)2(..............................2cos2sin
)1...(..............................2sin2cos
)1()1(
1
1
1
cxxxxI
generally
xxxx
xxxxI
xxxxI
Ixxxx
IxxxxI
getWe
IxxI
in
IxxI
equationinequationfromISubstitute
Another method to evaluate this integral and similar ones is to use the proper trigonometric
identities
Recall that:
)]cos()[cos(sinsin).....3(
)]cos()[cos(coscos)....2(
)]sin()[sin(cossin).....1(
21
21
21
bababa
bababa
bababa
Using the identity (2), we get:
cxx
cxx
cdxxx
dxxxI
3sin6
1sin2
1
]3sin3
1[sin2
1
]3cos[cos2
1
cos2cos
Home Quiz
1.Prove the identity(2) of the previously given trigonometric identities
2.Show that the two values arrived at for the integral of of this example are equivalent
Examples IIIUsing the method to find the integrals of trigonometric and inverse trigonometric functions that can not be found directly
A. ∫arcsinx dx , ∫arccosx dx , ∫arctanx dx , ∫arccotx dx, ∫arcsecx dx and ∫arccscx dx
B. ∫secnx dx and ∫cscnx dx , where n is an odd natural number greater than
1
Examples III - AExample 1
dxxI arcsin
cxxx
cx
xx
x
xdxxxI
xvandx
dxdu
dxdvandxu
Let
dxxI
2
21
2
21
2
2
1arcsin
)1(arcsin
1arcsin
1
arcsin
arcsin
21
Example 2
dxxI arctan
cxxx
x
xdxxxI
xvandx
dxdu
dxdvandxu
Let
dxxI
)1ln(arctan
1arctan
1
arctan
arctan
221
2
2
Example 3
dxxarcI sec
1sec
1sec
1
sec
sec
2
2
2
x
dxxarcx
xx
xdxxarcxI
xvandxx
dxdu
dxdvandxarcu
Let
dxxarcI
We find the last integral using the method of trigonometric substitution
cxx
cd
d
x
dxI
Thus
x
and
ddx
x
Letx
dxI
1ln
tanseclnsec
tan
tansec
1
,
tan1
tansec
sec
1
2
21
2
21
Substituting that back, we get:
cxxxarcxI
generalInxx
xdxxarcxI
1lnsec
,1
sec
2
2
Examples III - BExample 1
dxxI 3sec
dxxdxxxx
dxxxxx
dxxxxxI
xvanddxxxdu
dxxdvandxu
Let
dxxx
dxxI
secsectansec
)1(secsectansec
tansectansec
tantansec
secsec
secsec
sec
3
2
2
2
2
3
cxxxxI
generally
xxxxI
xxxxI
xxIxx
dxxdxxxxI
]tanseclntan[sec2
1
,
]tanseclntan[sec2
1
tanseclntansec2
tanseclntansec
secsectansec 3
Examples III - BExample 2
dxxI 5sec
dxxdxxxx
dxxxxx
dxxxxxI
xvanddxxxdu
dxxdvandxu
Let
dxxx
dxxI
353
233
233
3
23
23
5
sec3sec3tansec
)1(secsec3tansec
tansec3tansec
tantansec3
secsec
:
secsec
sec
cxxxxxx
I
Generally
xxxxxxI
xxxxxx
dxxxxI
dxxdxxxx
tanseclntansectansec
.
tanseclntansectansec
tanseclntan[sectansec
sec3tansec4
sec3sec3tansec
83
833
41
83
833
41
233
33
353
Questions
Do them as homework
Questions I
dxxx
dxxx
dxxx
dxex x
3 259
59
59
9
)72()4(
sin)3(
cos)2(
)1(5
Questions II
dxxx
dxxx
dxxx
n ln)3(
arctan)2(
arcsin)1(
Questions III
dxxecarcs
dxx
dxx
dxx
dxx
dxx
)3()6(
)3arccos()5(
)3arcsin()4(
)3arctan()3(
)cos(ln)2(
)sin(ln)1(
Questions IVReduction Formulas
dxxn
nxx
ndxx
dxxn
xdxx
dxxn
nxx
ndxx
dxxn
nxx
ndxx
nn
nn
n
nn
nnn
22
21
21
21
sec1
2tansec
1
1)3arcsin()4(
tan1
tantan)3(
cos1
sincos1
)cos(ln)2(
sin1
cossin1
sin)1(