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Section 7.2
Advanced Integration Techniques: Trigonometric Integrals
We will use the following identities quite often in this section; you would do well to memorize
them.
sin2 x = 1−cos(2x) cos2x = 1+cos(2x) (1)
2 2 2 2
cos(2x) = 1−2sin x cos(2x) = 2cos x−1
2 2 2 2
sec x = 1+tan x csc x = 1+cot x
When attempting to integrate a function built up from trigonometric functions, there are of-
ten many different possibilities for choosing an integration technique. For example, we can solve
∫ sinxcosxdx using the u-substitution u = cosx. The same substitution could be used to find
∫ tanxdx if we note that tanx = sinx. We can use integration by parts to solve ∫ sin(5x)cos(3x)dx.
cosx
However, there are many other trigonometric functions whose integrals can not be evaluated so eas-
ily. In this section, we will look at multiple techniques for handling integrals of several different
types of trig functions.
∫ m n
Integrals of the form sin xcos x
∫ m n
Tointegrate a function of the form sin xcos x, we will use one of the two following methods:
1. if both the powers m and n are even, rewrite both trig functions using the identities in (1)
2. Otherwise, we will rewrite the function so that only one power of sinx (or one power of cosx)
appears; this will allow us to make a helpful substitution:
(a) If m = 2k +1 is odd, then rewrite
m 2k+1 2k 2 k 2 k
sin x=sin x=(sinx)(sin x) = (sinx)(sin x) = (sinx)(1−cos x) ,
then use the u-substitution u = cosx.
(b) If n = 2k +1 is odd, then rewrite
n 2k+1 2k 2 k 2 k
cos x = cos x=(cosx)(cos x)=(cosx)(cos x) =(cosx)(1−sin x) ,
then use the u-substitution u = sinx.
Examples:
∫ 3
Find cos (2x)dx.
Since cos(2x) has an odd power, let’s rewrite
3 2 2
cos (2x) = cos(2x)cos (2x) = cos(2x)(1−sin (2x)).
Then ∫ ∫
3 2
cos (2x)dx = cos(2x)(1−sin (2x))dx.
1
Section 7.2
Wewillneedthesubstitution u = sin(2x) so that du = 2cos(2x)dx. Now we can finish the problem:
∫ 3 ∫ 2
cos (2x)dx = cos(2x)(1−sin (2x))dx
1 ∫ 2
=2 1−udu using the substitution u = sin(2x)
1( 1 3)
=2 u−3u +C
1 1 3
=2u−6u +C
1 1 3
=2sin(2x)− 6sin (2x)+C.
∫ 3 5
Find sin xcos xdx.
Since both trig functions have odd powers, we will rewrite one of them using the Pythagorean
identity. Let’s try
3 5 3 4
sin xcos x = sin xcos xcosx
=sin3x(cos2x)2cosx
3 2 2
=sin x(1−sin x) cosx.
As in the previous example, we can use a simple u-substitution to finish the problem. Set
u=sinx so that du = cosxdx. Then
∫ 3 2 2 ∫ 3 2 2
sin x(1−sin x) cosxdx = ∫ u (1−u ) du
3 2 4
=∫ u (1−2u +u )du
3 5 7
= u −2u +u du
1 4 2 6 1 8
=4u −6u +8u +C
1 4 1 6 1 8
=4sin x−3sin x+ 8sin x+C.
∫ 2
Find cos (2x)dx.
Since there are no odd powers in this function, we will rewrite cos2(2x) = 1+cos(4x) using the
2
2
Section 7.2
equation in (1). Then the integral calculation is fairly routine:
∫ cos2(2x)dx = ∫ 1+cos(4x)dx
∫ 2
=1 1+cos(4x)dx
2
=1(x+1sin(4x))+C using the substitution u = 4x
2 4
=1x+1sin(4x)+C.
2 8
∫ 2 4
Evaluate cos xsin xdx.
Since both the powers of cosx and sinx are even, we will write
cos2x = 1+cos(2x)
2
and ( )
2
4 2 2 1−cos(2x)
sin x = (sin x) = 2 .
Then
∫ 2 4 ∫ (1+cos(2x))(1−cos(2x))2
cos xsin xdx = 2 2 dx
∫ (1+cos(2x))(1−2cos(2x)+cos2(2x))
= 2 4 dx
=1∫ 1−2cos(2x)+cos2(2x)+cos(2x)−2cos2(2x)+cos3(2x)dx
8 ∫
1 2 3
=8( 1−cos(2x)−cos (2x)+cos (2x)dx )
1 1 ∫ 2 ∫ 3
=8 x−2sin2x− cos (2x)dx+ cos (2x) dx.
Wehave already showed that
∫ cos2(2x)dx = 1x+ 1 sin(4x)+C
2 8
and ∫
3 1 1 3
cos (2x)dx = 2 sin(2x)− 6 sin (2x)+C,
so finally we have
∫ 2 4 1( 1 1 1 1 1 3 )
cos xsin xdx = 8 x− 2 sin2x− 2x− 8 sin(4x)+ 2 sin(2x)− 6 sin (2x) +C.
3
Section 7.2
Integrating powers of tanx, secx, cscx, and cotx
Tointegratepowersoftheothertrigfunctions, wewilloftenneedtouseu-substitutionorintegration
by parts together with the pythagorean identities; if possible, we will need to take advantage of the
fact that d tanx = sec2x, d sec2x = secxtanx, d cscx = −cscxcotx, and d cotx = −csc2x.
dx dx dx dx
Example:
Evaluate ∫ csc4xdx.
4 2 2 2 2
Writing csc x = (csc x)(csc x) = (1+cot x)(csc x) is advantageous, as it will allow us to use
the substitution u = cotx:
∫ csc4xdx = ∫ (1+cot2x)(csc2x)dx
=−∫ (1+u2)du using u = cotx and −du = csc2xdx
=−u−1u3+C
3
=−cotx−1cot3x+C.
3
Eliminating square roots
If the function we wish to integrate involves the square root of some trigonometric function, we
may be able to eliminate the root by using the pythagorean identities or the identities from (1).
Examples:
Evaluate ∫ √cosy +1dy
The identity cos2x = 1+cos(2x) can help us here. Setting y = 2x, so that x = y, the identity
2 2
becomes cos2(y) = 1+cosy. We would like to replace the quantity cosy+1; solving for this quantity
2 2 2 y
in the above identity, we have cosy + 1 = 2cos (2). So we may rewrite the integral as
∫ √ ∫ √ 2(y)
cos(y) +1dx = 2cos 2 dy
∫ √ (y)
= 2cos 2 dy
√ (y)
=2 2sin 2 +C.
∫ √ 2
Find csc θ −1dθ.
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