Summary TRIGONOMETRIC INTEGRALS

Trigonometric Integrals

Suppose we have
∫sinm xcosn xdx
To find the integral, we must follow the following rules
(a)if the power of cosine is odd(n = 2k + 1), save one cosine factor
and use cos2 x = 1 - sin2 x to express the remaining in terms of
sine
for e.g ∫sinm xcos k 2 +1
xdx =∫sinm x(cos x)k cosxdx
2


k
= ∫sin x(1 - sin x) cosxdx
m 2



then substitute u = sinx {Use u - substitution}
(b)if the power of sine is odd(m = 2k + 1), save one sine factor and
use sin2 x = 1 - cos2 x to express the remaining factor in terms of
cosine
k
∫ sin2k + 1 xcosn xdx = ∫(sin2 x) cosn xsinxdx
= ∫(1 - cos x)k cosn xsinxdx
2


then substitute u = cosx

(c)if the powers of both and cosine are even, use the half angle
identities
1
sin2 x =
2
(1 - cos2x)
1
cos2 x =
2
(1 + cos2x)
1
sinxcosx = sin2x
2

Examples
Evaluate

, (a)∫cos3 xdx = ∫cos2 xcosxdx

= ∫(1 - sin x)cosxdx 2


let u = sinx
then du = cosxdx
du
therefore dx =
cosx
du
Thus = ∫(1 - u )cosx cosx
2




= ∫(1 - u )du
2



= ∫1du - ∫u du
2


1 3
= u- u +c
3

1 3
= sinx - sin x+c
3

We can also check our answer by differentiating it
d 1 d 1 d d
sinx - sin3 x + c = sinx - sin3 x + c
dx 3 dx 3 dx dx
1 2
= cosx - × 3sin xcosx + 0
3

= cosx - sin2 xcosx
= cosx (1 - sin2 x)
cosx (cos2 x) =
3
= cos x
Thus our answer is correct


(b)∫sin5 xcos2 xdx = ∫sin4 xcosxsinxdx

= ∫(sin2 x) cosxsinxdx2




= ∫(1 - cos x) cosxsinxdx
2 2



let u = cosx
then du = - sinxdx