02 Trigonometric Integrals - Handout.pdf

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Trig Integrals Trig Integrals Trig Integrals Exercises

Techniques of Integration–Trigonometric

Integrals

Mathematics 54–Elementary Analysis 2

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Trig Integrals Trig Integrals Trig Integrals Exercises or sin cos

Trigonometric Integrals

Integrals of the formR

sinmx dxorR

cosmx dx

Recall

Z

sin

x dx

= −

cos

x

+

C

Z

sin

2

x dx

=

Z

1

2

(1

−

cos 2

x

)

dx

=

1

2

x

−

1

(3)

Trig Integrals Trig Integrals Trig Integrals Exercises Rsinm x dx

orRcosm x dx R

sinm xcosn x dx

sinmx dxorR

cosmx dx

Example.

Consider

Z

sin

3

x dx

.

Note that

sin

3

x

=

sin

2

x

sin

x

=

(1

−

cos

2

x

) sin

x

=

sin

x

−

cos

2

x

sin

x

Thus,

Z

sin

3

x dx

=

Z

¡

sin

x

−

cos

2

x

sin

x

¢

dx

.

Let

u

=

cos

x

,

du

= −

sin

x dx

. Therefore,

Z

sin

3

x dx

= −

Z

¡

1

−

u

2

¢

du

= −

u

+

1

3

u

3

+

C

=

−

cos

x

+

1

3

cos

3

x

(4)

sinmx dxorR

cosmx dx

Z

sin

m

x dx

,

m

∈

N

m

is odd

split off a factor of sinx

express the rest of the factors in terms of cosx, using sin2x=1−cos2x

use the substitutionu=cosx, du= −sinx dx

m

is even

use the half-angle identity

sin2x₌1

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Trig Integrals Trig Integrals Trig Integrals Exercises Rsinm x dx

orRcosm x dx R

sinm xcosn x dx

Trigonometric Integrals

sinmx dxorR

cosmx dx

Z

cos

m

x dx

,

m

∈

N

m

is odd

split off a factor of cosx

express the rest of the factors in terms of sinx, using cos2x=1−sin2x

use the substitutionu=sinx, du=cosx dx

m

is even

use the half-angle identity

cos2x₌1

(6)

Example.

Evaluate

Z

cos

5

x dx

Z

cos5x dx = Z

cos4xcosx dx

= Z

¡ cos2x¢2

cosx dx₌ Z

¡

1₋sin2x¢2 cosx dx

= Z

¡

1−2sin2x+sin4x¢ cosx dx

Letu₌sinx, du₌cosx dx. Z

cos5x dx ₌ Z

¡

1−2u2₊u4¢ du

= u₋2 3u

3

+1₅u5₊C

(7)

Trig Integrals Trig Integrals Trig Integrals Exercises R

sinm x dxorR

cosm x dx Rsinm xcosn x dx

sinmxcosnx dx

Example.

Evaluate

Z

cos

3

x

sin

2

x dx

.

Z

cos

3

x

sin

2

x dx

=

R

cos

2

x

sin

2

x

cos

x dx

=

Z

¡

1

−

sin

2

x

¢

sin

2

x

cos

x dx

=

Z

sin

2

x

cos

x dx

−

Z

sin

4

x

cos

x dx

let

u

=

sin

x

du

=

cos

xdx

Z

cos

3

x

sin

2

x dx

=

Z

u

2

du

−

Z

u

4

du

=

1

3

u

3

−

1

5

u

5

(8)

Trig Integrals Trig Integrals Trig Integrals Exercises sin or cos

Trigonometric Integrals

sinmxcosnx dx

Z

sin

m

x

cos

n

x dx

m

is odd

split off a factor of sinx

express the rest of the factors in terms of cosx, using sin2x=1−cos2x

(9)

Trig Integrals Trig Integrals Trig Integrals Exercises R

sinm x dxorR

cosm x dx Rsinm xcosn x dx

sinmxcosnx dx

Z

sin

m

x

cos

n

x dx

n

is odd

split off a factor of cosx

express the rest of the factors in terms of sinx, using cos2x=1−sin2x

use the substitutionu=sinx, du=cosx dx

both

m

and

n

are even

use the half-angle identities

cos2x₌1

2(1+cos 2x) and sin 2_x

=1₂(1₋cos 2x)

use the rule for Z

(10)

Trig Integrals Trig Integrals Trig Integrals Exercises sin or cos

Example.

Evaluate

Z

sin

2

x

cos

4

x dx

.

Z

sin

2

x

cos

4

x dx

=

Z

sin

2

x

(cos

2

x

)

2

dx

=

Z

µ

1

−

cos 2

x

2

¶ µ

1

+

cos 2

x

2

¶

2

dx

=

Z

µ

1

−

cos 2

x

2

¶ µ

1

+

cos 2

x

2

¶

2

dx

=

1

8

Z

¡

1

+

cos 2

x

−

cos

2

x

−

cos

3

2

x

¢

dx

=

1

8

Z

·

1

+

cos 2

x

−

µ

1

+

cos 4

x

2

¶

−

(1

−

sin

2

x

) cos 2

x

¸

dx

=

1

8

·

x

+

sin 2

x

2

−

1

2

µ

x

+

sin 4

x

4

¶

−

1

2

µ

sin 2

x

−

sin

3

2

x

3

¶¸

(11)

Trig Integrals Trig Integrals Trig Integrals Exercises Rtanm x dx

or

Z

cotm x dx R

secn x dxorR

cscn x dx

Trigonometric Integrals

tanmx dxorR

cotmx dx

Example.

Evaluate

Z

tan

3

x dx

.

Z

tan

x

tan

2

x dx

=

Z

tan

x

¡

sec

2

x

−

1

¢

dx

=

Z

tan

x

sec

2

x dx

−

Z

tan

x dx

let

u

=

tan

x

,

du

=

sec

2

x dx

Z

tan

3

x dx

=

Z

u du

−

ln

|

sec

x| +

C

=

1

2

u

2

−

ln

|

sec

x

| +

C

=

1

2

¡

tan

2

x

¢

(12)

Trig Integrals Trig Integrals Trig Integrals Exercises tanm x dxor cotm x dx secn x dxor cscn x dx

tanmx dxorR

cotmx dx

Z

tan

m

x dx

split off a factor of tan

2

x

and write this as tan

2

x

=

sec

2

x

−

1

use the substitution

u

=

tan

x

,

du

=

sec

2

x dx

Z

cot

m

x dx

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Trig Integrals Trig Integrals Trig Integrals Exercises Rtanm x dx

or

Z

cotm x dx R

secn x dxorR

cscn x dx

Example.

Evaluate

Z

cot

4

3

x dx

.

Z

cot23xcot23x dx = Z

cot23x¡

csc23x−1¢ dx

= Z

¡

cot23xcsc23x₋cot23x¢ dx

= Z

¡

cot23xcsc23x−csc23x+1¢ dx

= Z

¡

cot23xcsc23x¢ dx₊1

3cot 3x+x+C letu₌cot 3x, du_{= −}3 csc23x dx

Z

cot43x dx ₌ −1 3

Z

u2du₊1

3cot 3x+x+C

= −₉1u3₊1

3cot 3x+x+C

= −1

9 cot

3₃_x

+1

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Trigonometric Integrals

secnx dxorR

cscnx dx

Example.

Evaluate

Z

csc

6

x dx

.

Z

csc

6

x dx

=

Z

(csc

2

x

)

2

csc

2

x dxdx

=

Z

¡

1

+

cot

2

x

¢

csc

2

xdx

=

Z

(1

+

2 cot

2

x

+

cot

4

x

) csc

2

x dx

let

u

=

cot

x

⇒

du

= −

csc

2

x dx

Z

csc

6

x dx

= −

R

(1

+

2

u

2

+

u

4

)

du

=

−

µ

cot

x

+

2 cot

3

x

3

+

cot

5

x

5

¶

(15)

Trig Integrals Trig Integrals Trig Integrals Exercises R

tanm x dxor

Z

cotm x dx Rsecn x dx

orRcscn x dx

Trigonometric Integrals

secnx dxorR

cscnx dx

Z

sec

n

xdx

n

is even

split off a factor of sec2x.

express the rest of the factors in terms of tanx, using sec2x₌1₊tan2x

use the substitutionu₌tanx, du₌sec2xdx.

Z

csc

n

xdx

n

is even

split off a factor of csc2x.

express the rest of the factors in terms of cotx, using csc2x₌1₊cot2x

(16)

Example.

Evaluate

Z

sec

3

x dx

.

Note that sec3x₌secxsec2x. By IBP,

u=secx , dv=sec2x dx du=secxtanx dx , v=tanx dx

Z

sec3x dx ₌secxtanx₋ Z

tanx(secxtanx)dx

=secxtanx₋ Z

tan2xsecx dx

=secxtanx₋ Z

(sec2x₋1) secx dx

Z

sec3x dx =secxtanx− Z

sec3x dx+ Z

secx dx

2 Z

sec3xdx ₌secxtanx₊ln|secx₊tanx_{| +}C

∴

Z

sec3xdx ₌1

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Trig Integrals Trig Integrals Trig Integrals Exercises R

tanm x dxor

Z

cotm x dx Rsecn x dx

orRcscn x dx

Trigonometric Integrals

secnx dxorR

cscnx dx

Z

sec

n

xdx

n

is odd

split off a factor of sec2x

use IBP withdv=sec2x dx anduto be the remaining factors

Z

csc

n

xdx

n

is odd

split off a factor of csc2x

(18)

Trig Integrals Trig Integrals Trig Integrals Exercises or sinmxcosnx dx, sinmxsinnx dxor cosmxcosnx dx

Trigonometric Integrals

tanmxsecnx dxorR

cotmxcscnx dx

Example.

Evaluate

Z

tan

3

x

sec

2

x dx

.

Z

tan

3

x

sec

2

x dx

=

Z

tan

2

x

sec

x

sec

x

tan

x dx

=

Z

¡

sec

2

x

−

1

¢

sec

x

sec

x

tan

x dx

=

Z

¡

sec

3

x

−

sec

x

¢

sec

x

tan

x dx

let

u

=

sec

x

,

du

=

sec

x

tan

x dx

Z

tan

3

x

sec

2

x dx

=

Z

¡

u

3

−u

¢

du

=

1

4

sec

4

x

−

1

2

sec

2

x

(19)

Trig Integrals Trig Integrals Trig Integrals Exercises Rtanm xsecn x dx

orRcotm xcscn x dx R

sinmxcosnx dx,R

sinmxsinnx dxorR

cosmxcosnx dx

Trigonometric Integrals

tanmxsecnx dxorR

cotmxcscnx dx

Z

tan

m

x

sec

n

x dx

m

is odd

split off a factor of secxtanx

express the rest of the factors in terms of secxusing the identity tan2x₌sec2x₋1

use the substitutionu₌secx, du₌secxtanx dx

Z

cot

m

x

csc

n

x dx

m

is odd

split off a factor of cscxcotx

express the rest of the factors in terms of cscxusing the identity cot2x₌csc2x₋1

(20)

Trigonometric Integrals

tanmxsecnx dxorR

cotmxcscnx dx

Z

tan

m

x

sec

n

x dx

n

is even

split off a factor of sec2x

express the rest of the factors in terms of tanxusing the identity sec2x₌1₊tan2x

use the substitutionu₌tanx, du₌sec2x dx

Z

cot

m

x

csc

n

x dx

n

is even

split off a factor of csc2x

express the rest of the factors in terms of cotxusing the identity csc2x₌1₊cot2x

(21)

sinmxcosnx dx,R

sinmxsinnx dxorR

cosmxcosnx dx

Example.

Evaluate

Z

cot

2

x

csc

x dx

.

Z

cot

2

x

csc

x dx

=

Z

(csc

2

x

−

1) csc

x dx

=

Z

(csc

3

x

−

csc

x

)

dx

=

Z

csc

3

x dx

−

ln

|

csc

x

−

cot

x

|

Exercise:

Z

csc3x dx_{= −}1

2cscxcotx+ 1

2ln|cscx−cotx| +C

=

−

1

2

csc

x

cot

x

−

1

(22)

Example.

Evaluate

Z

p

tan

x

sec

4

x dx

.

Z

p

tan

x

sec

4

x dx

=

Z

p

tan

x

sec

2

x

sec

2

x dx

=

Z

p

tan

x

(1

+

tan

2

x

) sec

2

x dx

=

Z

³

p

tan

x

+

p

tan

5

x

´

sec

2

x dx

=

2

3

p

tan

3

x

+

2

7

p

(23)

sinmxcosnx dx,R

sinmxsinnx dxorR

cosmxcosnx dx

Trigonometric Integrals

tanmxsecnx dxorR

cotmxcscnx dx

Z

tan

m

x

sec

n

x dx

m

is even and

n

is odd

express the even power of tanxin terms of secxusing the identity tan2x₌sec2x₋1

use the rule for Z

secmx dx

Z

cot

m

x

csc

n

x dx

m

is even and

n

is odd

express the even power of cotxin terms of cscxusing the identity cot2x=csc2x−1

use the rule for Z

(24)

Trig Integrals Trig Integrals Trig Integrals Exercises tan sec or cot csc sinmxcosnx dx, sinmxsinnx dxor cosmxcosnx dx

Trigonometric Integrals

F. Integrals of the formR

sinmxcosnxdx,R

sinmxsinnxdxorR

cosmxcosnxdx

Recall. Product to Sum Formula

sin

mx

cos

nx

=

1

2

[sin(

m

+

n

)

x

+

sin(

m

−

n

)

x

],

sin

mx

sin

nx

=

−

1

2

[cos(

m

+

n

)

x

−

cos(

m

−

n

)

x

],

cos

mx

cos

nx

=

1

(25)

Trig Integrals Trig Integrals Trig Integrals Exercises R

tanm xsecn x dxorR

cotm xcscn x dx R

sinmxcosnx dx,R

sinmxsinnx dxorR

cosmxcosnx dx

Example.

Evaluate

Z

cos 3

x

cos 5

x dx

.

Z

cos 3

x

cos 5

x dx

=

1

2

Z

(cos(3

x

+

5

x

)

+

cos(3

x

−

5

x

))

dx

=

1

2

Z

(cos 8

x

+

cos 2

x

)

dx

=

1

2

µ

1

8

sin 8

x

+

1

2

sin 2

x

¶

+

C

=

1

16

sin 8

x

+

1

(26)

Exercises

Evaluate the following integrals.

1

Z

1 0

sin

2

π

x

cos

2

π

x dx

2

Z

cos

3

x

p

sin

x

dx

3

Z

csc

4

x

cot

2

x

dx

4

Z

cos 4

x

cos 3

x dx

5

Z

tan

3

(ln

x

) sec

8

(ln

x

)