Trigonometric Integrals: Transforming Complicated Integrals using Identities - Prof. Qingl, Study notes of Calculus

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8.4 Trigonometric Integrals

Idea: the general idea is to use identities to transform complicated trigono-

metric integrals into integrals that are easier to work with.

Products of powers of sines and cosines

Evaluate (^) ∫

sin

m x cos

n xdx

where m, n are integers.

Key identity:

sin

2 x + cos

2 x = 1.

Case 1: If m (i.e. the power of sin x) is odd, then use the substitution

u = cos x.

sin

m x cos

n xdx

sin

2 k+ x cos

n xdx

sin

2 k x cos

n x sin xdx

1 − cos

2 x

)k

cos

n x sin xdx

1 − u

2

)k

u

n du by setting u = cos x

So, the integrand becomes a polynomial or a rational function.

Example: ∫ cos

2 x

sin x

dx

cos

2 x

sin x

dx

cos

2 x

sin

2 x

sin xdx

u

2

1 − u

2

du by setting u = cos x

∫ [

1 − u^2

]

du

∫ [

1 − u

1 + u

]

du

= u +

ln | 1 − u| −

ln |1 + u| + C

= u +

ln

1 − u

1 + u

+ C

= u +

ln

1 − u

1 + u

+ C

= cos x +

ln

1 − cos x

1 + cos x

+ C

= cos x +

ln

(1 − cos x)

2

1 − cos^2 x

+ C

= cos x +

ln

(1 − cos x)

2

sin

2 x

+ C

= cos x + ln

1 − cos x

sin x

+ C

More generally, ∫

sin

2 k+ (x)f (cos x) dx

sin

2 k (x)f (cos x) sin xdx

1 − cos

2 x

)k

f (cos x) sin xdx

1 − u

2

)k

f (u) du by setting u = cos x

Example: ∫

sin

3 (x)

3 cos

5 x + 4 cos x + 4

1 − 2 cos^2 x + 2 cos x

dx

Let u = cos x, then ∫

sin

3 x

3 cos

5 x + 4 cos x + 4

1 − 2 cos^2 x + 2 cos x

dx

1 − u

2

3 u

5

• 4u + 4

1 − 2 u

2

• 2u

du

It becomes an integral of a rational function. So, we may use the method of

partial fractions to solve it.

Evaluate (^) ∫

sin

m x cos

n xdx

Case 2: If n (i.e. the power of cos x) is odd, then use the substitution

u = sin x.

sin

m x cos

n xdx

sin

m x cos

2 k+ xdx

sin

m x

1 − sin

2 x

)k

cos xdx

u

m

1 − u

2

)k

du by setting u = sin x.

Example: (^) ∫

cos

3 (x)e

sin x dx

Let u = sin x, then ∫

cos

3 xe

sin x dx

1 − u

2

e

u du

1 − u

2

e

u

2 ue

u du, by using integration by parts

1 − u

2

e

u

• 2ue

u − 2 e

u

• C

−u

2

• 2u − 1

e

u

• C

= − (u − 1)

2 e

u

• C

= − (sin x − 1)

2 e

sin x

• C

Case 3: If both m and n are even, that is m = 2k, n = 2l. Then, we

substitute

sin

2 x =

1 − cos 2x

and cos

2 x =

1 + cos 2x

to reduce the integrand to one in lower powers of cos 2x.

sin

2 k x cos

2 l xdx

1 − cos 2x

k (

1 + cos 2x

l dx

Example: (^) ∫

sin

4 x cos

2 xdx

sin

4 x cos

2 xdx

1 − cos 2x

1 + cos 2x

dx

1 − 2 cos 2x + cos

2 2 x

(1 + cos 2x) dx

1 − cos 2x − cos

2 2 x + cos

3 2 x

dx

sin

4 x cos

2 xdx

1 − cos 2x

1 + cos 2x

dx

1 − 2 cos 2x + cos

2 2 x

(1 + cos 2x) dx

1 − cos 2x − cos

2 2 x + cos

3 2 x

dx

cos

2 2 xdx

1 + cos 4x

dx

x

sin 4x

+ C

Therefore, ∫

sin

4 x cos

2 xdx

1 − cos 2x − cos

2 2 x + cos

3 2 x

dx

[

x −

sin 2x

x

sin 4x

sin (2x)

sin

3 (2x)

)]

+ C

[

x

sin 4x

sin

3 (2x)

]

+ C

[

x −

sin 4x

sin

3 (2x)

]

+ C

Eliminating square roots

Example: ∫ π 4

0

1 + cos 4xdx