Integration Techniques Summary

7/30/2019 Integration Techniques Summary

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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


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[ ])(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 =+ .


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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


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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 ++


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(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


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=

+

+

+++ 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)

Integration Techniques Summary

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