7/30/2019 Integration Techniques Summary
1/6
Integration Techniques Summary
1. List of basic formulas:
Function Integrated Result
nax
1
1
+
+
n
ax n
nbax )( +
( ) )1(
)( 1
++ +
na
bax n
[ ][ ]
n
xfxf )()(' [ ]1
)(
1
+
+
nxf
n
[ ]
[ ] 22 )(
)('
xfa
xf
+
a
xf
a
)(tan
1 1
[ ]
[ ] 22)(
)('
axf
xf
axfaxf
a +
)(
)(ln
2
1
[ ]
[ ] 22 )(
)('
xfa
xf
)()(
ln2
1
xfa
xfa
a +
[ ]
[ ] 22 )(
)('
xfa
xf
a
xf )(sin 1
)()(' xfaxf )(
ln1 xfaa
)()(' xfexf )(xfe
)(
)('
xf
xf |)(|ln xf
Function Integrated Result
7/30/2019 Integration Techniques Summary
2/6
[ ])(cos)(' xfxf [ ])(sin xf
[ ])(sin)(' xfxf [ ])(cos xf
[ ])(sec)(' 2 xfxf [ ])(tan xf
[ ])(cos)(' 2 xfecxf [ ])(cot xf
[ ] [ ])(cot)(cos)(' xfxfecxf [ ])(cos xfec
xtan |cos|ln x
xcot |sin|ln x
xsec |tansec|ln xx +
ecxcos |cotcos|ln xecx
2. Miscellaneous trigonometric integrals:
a. xdxn
sin and xdxn
cos :
If n is odd, separate a single xsin or xcos from the original function, and use the
identity 1cossin 22 =+ xx .
Example: ( )( ) == dxxxdxxxxdx ))(sincos1(sinsinsin223
= Cx
xdxxxx ++= 3cos
cossincossin3
2 (shown)
If n is even, use the double angle formula of either 1cos22cos 2 = xx orxx 2sin212cos = for conversion.
Example: Cxxdxx
xdx ++=+
= 21
2sin4
1
2
12coscos2 (shown)
b. xdxntan :
Separate x2tan from the original function, and use the identity xx 22 sectan1 =+ .
7/30/2019 Integration Techniques Summary
3/6
Example: ( )( ) ( )( )dxxxdxxxxdx ==32325 tan1sectantantan
= dxxxx 332 tantansec
= dxxxxx
)(tantantansec 232
= ( ) dxxxxx )(tan1sectansec232
= dxxxxxx + tantansectansec232
= Cxxx + |cos|lntan2
1tan
4
1 24(shown)
c. dxnxmx )sin()sin( or dxnxmx )cos()sin( or dxnxmx )cos()cos( :Use one of the 3 identities below to transform the product to a sum or difference:
(i) ( )[ ] ( )[ ]{ }xnmxnmnxmx ++= sinsin21
)cos()sin(
(ii) ( )[ ] ( )[ ]{ }xnmxnmnxmx += coscos2
1)sin()sin(
(iii) ( )[ ] ( )[ ]{ }xnmxnmnxmx ++= coscos2
1)cos()cos(
Example: ++=+= Cxxdxxxxdxx 2sin41
4sin8
12cos4cos
2
1cos3cos (shown)
3. Integration via substitution:
Generally, to find an integral by means of a substitution ),(ufx =
(i) Differentiate x wrtu to arrive at duufdxufdu
dx)(')(' ==
(ii) Subsequently replace all x by )(uf and dx by duuf )('
(iii) Perform the integration and remember to convert the result back to x
for an indefinite integral ( not needed for definite integrals)
Example: Evaluate + 12 xx
dxby considering the substitution tan
2
3
2
1=x .
dx
xxx
dx
+
=+
22
2
)2
3()
2
1(
1
1
Using the substitution tan2
3
2
1=x , the integral becomes
7/30/2019 Integration Techniques Summary
4/6
1
3
4tan
2
3
4
3
2secd
=
d)sec2
3(
sec2
3
1
= d)(sec
=C++ |tansec|ln
=ln|)
2
1(
3
2])
2
1(
4
3[
3
2| 2
++xx
+C (shown)
2)2
1(
4
3+ x
2
1x
2
3
4. Integration by parts:
= vdxuuvdxuv '' where u is a function which can be differentiated and v isa function that can be easily reduced via integration. Note that integration by parts
is only feasible if out of the product of two functions, at leastone is directly
integrable.
Example:
dxxx )(sin 21
= dxxx
xx
x)2)(
1
1(
2)(sin
2 4
221
2
= dxx
xx
x)
1()(sin
2 4
321
2
= dxx
xx
x)
1
4(
4
1)(sin
2 4
321
2
= Cxxx
+ 4212
1)2(4
1)(sin
2
= Cxxx + 4212
12
1)(sin2
(shown)
5. Integrals of the form dxcbxax
dcx ++
+2
(i) Rewrite dxcbxax
dcx ++
+2
as dxcbxax
Q
cbxax
baxP
cbxax
QbaxP
+++
+++
=++++
222)2()2(
,
where bax +2 is clearly the derivative of cbxax ++2 .
(ii) +++
cbxax
baxP2
)2(gives ||ln
2 cbxaxP ++
7/30/2019 Integration Techniques Summary
5/6
(iii) dxcbxax
Q ++2 can be easily integrated via completing the square method for
the denominator.
Example: dxxxdxxx
x
dxxx
x
+++++
=++
1
1
2
3
1
12
2
1
1
1222
=
+
+
++ dx
x
xx22
2
2
3
2
1
1
2
3|1|ln
2
1
= C
x
xx +
+
++
2
3
2
1
tan
23
1
2
3|1|ln
2
1 12
= Cx
xx +
+++
3
12tan3|1|ln
2
1 12(shown)
6. Integrals of the form dxcbxax
edxcx ++
++2
2
(i) Rewrite( )
dxcbxax
RbaxQcbxaxPdxcbxax
edxcx ++
+++++=
++++
2
2
2
2 )2(
= dxcbxax
R
cbxax
baxQP
+++
+++
+ 22)2(
dxcbxax
R
cbxax
baxQPx
+++
+++
+= 22)2(
where bax +2 is clearly the derivative of cbxax ++2 .
(ii) +++
cbxax
baxQ2
)2(gives ||ln
2 cbxaxQ ++
(iii) dxcbxax
R ++2 can be easily integrated via completing the square method for
the denominator.
Example:( )
++++++
=++++
dxxx
xxxdx
xx
xx
1
2)12(12
1
1422
2
2
2
++++ ++=dx
xxxxxx
12
1122
22
7/30/2019 Integration Techniques Summary
6/6
=
+
+
+++ dx
x
xxx22
2
2
3
2
1
12|1|ln2
= 2|1|ln22 +++ xxx C
x
+
+
2
3
2
1
tan
2
3
1 1
= Cx
xxx +
++++
3
12tan
3
4|1|ln2 12 (shown)
Top Related