Techniques of Integration–Trigonometric Integrals

Techniques of Integration–TrigonometricIntegrals

Mathematics 54–Elementary Analysis 2

Institute of MathematicsUniversity of the Philippines-Diliman

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Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx∫sinm x dx or

∫cosm x dx

Recall∫sinx dx =−cosx+C∫sin2 x dx =

∫1

2(1−cos2x) dx = 1

2x− 1

4sin2x+C

2 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx∫sinm x dx or

∫cosm x dx

Consider∫

sin3 x dx.

Note that

sin3 x = sin2 x sinx

= (1−cos2 x) sinx

= sinx−cos2 x sinx

Thus,∫

sin3 x dx =∫ (

sinx−cos2 x sinx)

dx.

Let u = cosx, du =−sinx dx. Therefore,∫sin3 x dx =−

∫ (1−u2) du =−u+ 1

3u3 +C =−cosx+ 1

3cos3 x+C

3 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

∫sinm x dx, m ∈N

m is oddsplit off a factor of sinx

express the rest of the factors in terms of cosx, usingsin2 x = 1−cos2 x

use the substitution u = cosx, du =−sinx dx

m is evenuse the half-angle identity

sin2 x = 1

2(1−cos2x)

4 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

∫cosm x dx, m ∈N

m is oddsplit off a factor of cosx

express the rest of the factors in terms of sinx, usingcos2 x = 1− sin2 x

use the substitution u = sinx, du = cosx dx

m is evenuse the half-angle identity

cos2 x = 1

2(1+cos2x)

5 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

Example 1.

Evaluate∫

cos5 x dx

∫cos5 x dx =

∫cos4 x cosx dx

=∫ (

cos2 x)2

cosx dx =∫ (

1− sin2 x)2

cosx dx

=∫ (

1−2sin2 x+ sin4 x)

cosx dx

Let u = sinx, du = cosx dx.∫cos5 x dx =

∫ (1−2u2 +u4) du

= u− 2

3u3 + 1

5u5 +C

= sinx− 2

3sin3 x+ 1

5sin5 x+C

6 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

Trigonometric IntegralsA. Integrals of the form

∫sinm x cosn x dx

Example 2.

Evaluate∫

cos2 x sin3 x dx.

∫cos2 x sin3 x dx = ∫

cos2 x sin2 x sinx dx

=∫

cos2 x(1−cos2 x

)sinx dx

=∫

cos2 x sinx dx−∫

cos4 x sinx dx

let u = cosx du =−sinxdx

=−∫

u2 du+∫

u4 du

=−1

3u3 + 1

5u5 +C =−1

3cos3 x+ 1

5cos5 x+C

7 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

∫cosm x sinn x dx

n is oddsplit off a factor of sinx

express the rest of the factors in terms of cosx, usingsin2 x = 1−cos2 x

use the substitution u = cosx, du =−sinx dx

8 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

Example 3.

Evaluate∫

cos3 x sin2 x dx.

∫cos3 x sin2 x dx = ∫

cos2 x sin2 x cosx dx

=∫ (

1− sin2 x)

sin2 x cosx dx

=∫

sin2 x cosx dx−∫

sin4 x cosx dx

let u = sinx du = cosxdx

=∫

u2 du−∫

u4 du

= 1

3u3 − 1

5u5 +C = 1

3sin3 x− 1

5sin5 x+C

9 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

∫cosm x sinn x dx

m is oddsplit off a factor of cosx

express the rest of the factors in terms of sinx, usingcos2 x = 1− sin2 x

use the substitution u = sinx, du = cosx dx

both m and n are evenuse the half-angle identities

cos2 x = 1

2(1+cos2x) and sin2 x = 1

2(1−cos2x)

use the rule for∫

cosm x dx

10 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx∫cosm x sinn x dx

Example 4.

Evaluate∫

sin2 x cos4 x dx.

∫sin2 x cos4 x dx =

∫sin2 x (cos2 x)2dx

=∫ (

1−cos2x

2

)(1+cos2x

2

)2

dx

=∫ (

1−cos2x

2

)(1+cos2x

2

)2

dx

= 1

8

∫ (1+cos2x−cos2 2x−cos3 2x

)dx

= 1

8

∫ [1+cos2x−

(1+cos4x

2

)− (1− sin2 2x)cos2x

]dx

= 1

8

[x+ sin2x

2− 1

2

(x+ sin4x

4

)− 1

2

(sin2x− sin3 2x

3

)]+C

11 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

Example.

Evaluate∫

tan3 2x dx.

∫tan2x tan2 2x dx =

∫tan2x

(sec2 2x−1

)dx

=∫

tan2x sec2 2x dx−∫

tan2x dx

let u = tan2x, du = 2sec2 2x dx

= 1

2

∫udu− 1

2ln |sec2x|+C

= 1

4u2 − 1

2ln |sec2x|+C

= 1

4

(tan2 2x

)− 1

2ln |sec2x|+C

12 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

∫tanm x dx, m ∈N,m ≥ 2

split off a factor of tan2 x and write this as tan2 x = sec2 x−1

use the substitution u = tanx, du = sec2 x dx

∫cotm x dx, m ∈N,m ≥ 2

split off a factor of cot2 x and write this as cot2 x = csc2 x−1

use the substitution u = cotx, du =−csc2 x dx

13 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

Example.

Evaluate∫

cot4 3x dx.

∫cot2 3x cot2 3x dx =

∫cot2 3x

(csc2 3x−1

)dx

=∫ (

cot2 3x csc2 3x−cot2 3x)

dx

=∫ (

cot2 3x csc2 3x−csc2 3x+1)

dx

=∫ (

cot2 3x csc2 3x)

dx+ 1

3cot3x+x+C

let u = cot3x, du =−3csc2 3x dx

= 1

3

∫u2 + 1

3cot3x+x+C

= 1

9u3 + 1

3cot3x+x+C

= 1

9cot3 3x+ 1

3cot3x+x+C

14 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx∫secn x dx or

∫cscn x dx

Example.

Evaluate∫

sec4 x dx.

∫sec4 x dx =

∫ (sec2 x

)(sec2 x

)dx

=∫ (

1+ tan2 x)

sec2 xdx

=∫

sec2 x dx+∫

tan2 x sec2 xdx

= tanx+∫

tan2 x sec2 xdx

let u = tanx, du = sec2 x dx

= tanx+∫

u2 du

= tanx+ tan3 x

3+C

15 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

∫secn xdx

n is evensplit off a factor of sec2 x.express the rest of the factors in terms of tanx, usingsec2 x = 1+ tan2 xuse the substitution u = tanx, du = sec2 xdx.

∫cscn xdx

n is evensplit off a factor of csc2 x.express the rest of the factors in terms of cotx, usingcsc2 x = 1+cot2 xuse the substitution u = cotx, du =−csc2 xdx

16 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx∫secn x dx or

∫cscn x dx

Example.

Evaluate∫

sec3 x dx.

Note that sec3 x = secx sec2 x. By IBP,

Let u = secx , dv = sec2 x dxdu = secx tanx dx , v = tanx dx∫

sec3 x dx = secx tanx−∫

tanx(secx tanx) dx

= secx tanx−∫

secx tan2 x dx

= secx tanx−∫

(sec2 x−1)secx dx

= secx tanx−∫

sec3 x dx+∫

secx dx

2∫

sec3 xdx = secx tanx+ ln |secx+ tanx|+C

∴∫

sec3 xdx = 1

2(secx tanx+ ln |secx+ tanx|)+C

17 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

∫secn xdx

n is oddsplit off a factor of sec2 xsuse IBP with dv = sec2 x dx and u to be the remaining factors

∫cscn xdx

n is oddsplit off a factor of csc2 xuse IBP, with dv = csc2 x dx and u to be the remaining factors

18 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

Example.

Evaluate∫

tan3 4x sec4 4x dx.

∫tan3 4x sec4 4x dx =

∫tan2 4x sec3 4x sec4x tan4x dx

=∫ (

sec2 4x −1)

sec3 4x sec4x tan4x dx

=∫ (

sec5 4x− sec3 3x)

sec4x tan4x dx

let u = sec4x, du = 4sec4x tan4x dx

= 1

4

∫ (u5 −u3) du

= 1

4

(u6

6− u4

4

)+C

= 1

24sec6 4x− 1

16sec4 4x+C

19 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

∫tanm x secn x dx

m is oddsplit off a factor of secx tanxexpress the rest of the factors in terms of secx using the identitytan2 x = sec2 x−1use the substitution u = secx, du = secx tanx dx

∫cotm x cscn x dx

m is oddsplit off a factor of cscx cotxexpress the rest of the factors in terms of cscx using the identitycot2 x = csc2 x−1use the substitution u = cscx, du =−secx tanx dx

20 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

Example.

Evaluate∫

tan3 4x sec4 4x dx.

∫tan3 4x sec4 4x dx =

∫tan3 4x sec2 4x sec2 4x dx

=∫

tan3 4x(1+ tan2 4x

)sec2 4x dx

=∫

tan3 4x sec2 4x dx+∫

tan5 4x sec2 4x dx

let u = tan4x, du = 4sec2 4x dx

= 1

4

∫ (u3 +u5) du

= 1

4

(u4

4+ u6

6

)+C

= 1

16tan4 4x+ 1

24tan6 4x+C

21 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

∫tanm x secn x dx

n is evensplit off a factor of sec2 xexpress the rest of the factors in terms of tanx using the identitysec2 x = 1+ tan2 xuse the substitution u = tanx, du = sec2 x dx

∫tanm x secn x dx

n is evensplit off a factor of csc2 xexpress the rest of the factors in terms of cotx using the identitycsc2 x = 1+cot2 xuse the substitution u = cotx, du =−csc2 x dx

22 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx∫tanm x secn x dx or

∫cotm x cscn x dx

Example.

Evaluate∫

cot2 x cscx dx.

∫cot2 x cscx dx =

∫(csc2 x−1)cscx dx

=∫

(csc3 x−cscx) dx

=∫

csc3 x dx− ln |cscx−cotx|

Recall:∫

csc3 x dx =− 1

2cscx cotx+ 1

2ln |cscx− tanx|+C

=−1

2cscx cotx− 1

2ln |cscx−cotx|+C

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Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

∫tanm x secn x dx

m is even and n is oddexpress the even power of tanx in terms of secx using theidentity tan2 x = sec2 x−1

use the rule for∫

secm x dx

∫cotm x cscn x dx

m is even and n is oddexpress the even power of cotx in terms of cscx using theidentity cot2 x = csc2 x−1

use the rule for∫

cscm x dx

24 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

Recall

sinmx cosnx = 1

2[sin(m+n)x+ sin(m−n)x],

sinmx sinnx = −1

2[cos(m+n)x−cos(m−n)x],

cosmx cosnx = 1

2[cos(m+n)x+cos(m−n)x].

25 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

Example.

Evaluate∫

cos3x cos5x dx.

∫cos3x cos5x dx = 1

2

∫(cos(3x+5x)+cos(3x−5x)) dx

= 1

2

∫(cos8x+cos2x) dx

= 1

2

(1

8sin8x+ 1

2sin2x

)+C

= 1

16sin8x+ 1

4sin2x+C

26 / 27

Techniques of Integration–Trigonometric Integrals∫

secn x dx or∫

cscn x dx

Example.

Evaluate∫

sin7x cos3x dx.

∫sin7x cos3x dx = 1

2

∫(sin(7x+3x)+ sin(7x−3x)) dx

= 1

2

∫(sin10x+ sin4x) dx

= 1

2

(− 1

10cos10x− 1

4cos4x

)+C

= − 1

20cos8x− 1

8cos4x+C

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