Download Advanced Integration Techniques: Trigonometric

Section 7.2
Advanced Integration Techniques: Trigonometric Integrals
When attempting to evaluate integrals of trig functions, it often helps to rewrite the function of
interest using an identity. Thus we will use the following identities quite often in this section; you
would do well to memorize them.
sin2 (αx) =
1−cos(2αx)
2
2
1+cos(2αx)
2
2
2 cos x − 1
cos2 (αx) =
cos(2x) = 1 − 2 sin x
cos(2x) =
sec2 x
csc2 x = 1 + cot2 x
=1+
tan2 x
(1)
There are many different possibilities for choosing an integration technique for an integral
involving trigonometric functions. For example, we can solve
Z
sin x cos x dx
using the u-substitution u = cos x. The same substitution could be used to find
Z
tan x dx
if we note that tan x =
sin x
cos x .
We can use integration by parts to solve
Z
sin(5x) cos(3x) dx.
However, there are many other trigonometric functions whose integrals can not be evaluated so
easily. In this section, we will look at multiple techniques for handling integrals of several different
types of trig functions.
Z
Integrals of the form
sinm x cosn x
To integrate a function of the form
Z
sinm x cosn x dx,
which is a product of (positive integer) powers of sin x and cos 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. If at least one of the powers is odd, we will rewrite the original function so that only one power
of sin x (or one power of cos x) appears; this will allow us to make a helpful substitution:
(a) If m = 2k + 1 is odd, then rewrite
sinm x = sin2k+1 x = (sin x)(sin2k x) = (sin x)(sin2 x)k = (sin x)(1 − cos2 x)k ,
and use the u-substitution u = cos x.
1
Section 7.2
(b) If n = 2k + 1 is odd, then rewrite
cosn x = cos2k+1 x = (cos x)(cos2k x) = (cos x)(cos2 x)k = (cos x)(1 − sin2 x)k ,
and use the u-substitution u = sin x.
Note: The general idea behind this technique actually works for any integral of the form
Z
sinm (βx) cosn (βx),
where β is any real number.
Z
Example. Find
cos3 (2x) dx.
Since cos(2x) has an odd power, let’s rewrite
cos3 (2x) = cos(2x) cos2 (2x) = cos(2x)(1 − sin2 (2x)).
Then
Z
cos3 (2x)dx =
Z
cos(2x)(1 − sin2 (2x)) dx.
We will need the substitution u = sin(2x) so that du = 2 cos(2x) dx. Now we can finish the
problem:
Z
Z
cos3 (2x) dx = cos(2x)(1 − sin2 (2x)) dx
Z
1
=
1 − u2 du
using the substitution u = sin(2x)
2
!
1
1 3
=
u− u +C
2
3
1
1
= u − u3 + C
2
6
1
1
= sin(2x) − sin3 (2x) + C.
2
6
Z
Example. Find
sin3 x cos5 xdx.
Since both trig functions have odd powers, we will rewrite one of them using the Pythagorean
identity. Let’s try
sin3 x cos5 x = sin3 x cos4 x cos x
= sin3 x(cos2 x)2 cos x
= sin3 x(1 − sin2 x)2 cos x.
2
Section 7.2
As in the previous example, we can use a simple u-substitution to finish the problem. Set
u = sin x so that du = cos x dx.
Then
Z
sin3 x(1 − sin2 x)2 cos x dx =
Z
u3 (1 − u2 )2 du
Z
u3 (1 − 2u2 + u4 ) du
Z
u3 − 2u5 + u7 du
=
=
1
2
1
= u4 − u6 + u8 + C
4
6
8
1
1
1
= sin4 x − sin6 x + sin8 x + C.
4
3
8
Z
Example. Find
cos2 (2x)dx.
Since there are no odd powers in this function, we will rewrite the integrand as
cos2 (2x) =
1 + cos(4x)
2
using the equation in (1). Then the integral calculation is fairly routine:
Z
Z
1 + cos(4x)
2
cos (2x) dx =
dx
2
Z
1
=
1 + cos(4x) dx
2
1
1
using the substitution u = 4x
= (x + sin(4x)) + C
2
4
1
1
= x + sin(4x) + C.
2
8
Z
Example. Evaluate
cos2 x sin4 x dx.
Since both the powers of cos x and sin x are even, we will write
cos2 x =
1 + cos(2x)
2
and
1 − cos(2x)
sin4 x = (sin2 x)2 =
2
3
!2
.
Section 7.2
Then
Z
cos2 x sin4 x dx =
Z
Z
=
1 + cos(2x)
2
!
1 + cos(2x)
2
!
1 − cos(2x)
2
!2
dx
1 − 2 cos(2x) + cos2 (2x)
4
!
dx
Z
1
1 − 2 cos(2x) + cos2 (2x) + cos(2x) − 2 cos2 (2x) + cos3 (2x) dx
8
Z
1
1 − cos(2x) − cos2 (2x) + cos3 (2x) dx
=
8
!
Z
Z
1
1
2
3
=
x − sin 2x − cos (2x) dx + cos (2x) dx.
8
2
=
We have already showed that
Z
and
Z
1
1
cos2 (2x) dx = x + sin(4x) + C
2
8
cos3 (2x) dx =
1
1
sin(2x) − sin3 (2x) + C,
2
6
so finally we have
Z
1
1
1
1
1
1
cos2 x sin4 x dx =
x − sin 2x − x − sin(4x) + sin(2x) − sin3 (2x)
8
2
2
8
2
6
!
+ C.
Integrating powers of tan x, sec x, csc x, and cot x
To integrate powers of the other trig functions, we will often need to use u-substitution or integration
by parts together with the pythagorean identities; if possible, we will need to take advantage of the
fact that
d
d
tan x = sec2 x,
sec2 x = sec x tan x,
dx
dx
d
d
csc x = − csc x cot x, and
cot x = − csc2 x.
dx
dx
Z
Example. Evaluate
csc4 xdx.
Rewriting
csc4 x = (csc2 x)(csc2 x)
= (1 + cot2 x)(csc2 x)
4
Section 7.2
is advantageous, as it will allow us to use the substitution u = cot x:
Z
csc4 x dx =
Z
(1 + cot2 x)(csc2 x) dx
Z
= − (1 + u2 ) du
using u = cot x and − du = csc2 x dx
1
= −u − u3 + C
3
1
= − cot x − cot3 x + 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).
Z p
Example. Evaluate
cos y + 1 dy.
The identity
cos2 (αx) =
1 + cos(2αx)
2
can help us here. We have
y
1 + cos y
cos
=
.
2
2
2
We would like to replace thequantity cos y + 1; solving for this expression in the above identity,
we have cos y + 1 = 2 cos2 y2 . So we may rewrite the integral as
Z p
Z s
y
cos(y) + 1 dy =
2 cos2
dy
2
Z √
y
=
2 cos
dy
2
√
y
= 2 2 sin
+ C.
2
Example. Find
Z p
csc2 θ − 1 dθ.
5
Section 7.2
Since csc2 θ − 1 = cot2 θ, let’s rewrite
Z p
Z √
csc2 θ − 1 dθ =
cot2 θ dθ
Z
= cot θ dθ.
cos θ
Since cot θ =
, we can integrate the function using a substitution; setting u = sin θ so that
sin θ
du = cos θ dθ, we have
Z p
Z
2
csc θ − 1 dθ =
Z
=
Z
=
cot θ dθ
cos θ
dθ
sin θ
1
du
u
= ln u + C
= ln | sin θ| + C.
A brief aside
Z
We have not yet learned how to evaluate sec x dx, and as we will need to know this integral
in future sections, let’s go ahead and compute it. It turns out that the best way to evaluate the
integral is by using Mathemagic: note that sec x can be rewritten as
sec x = sec x
sec x + tan x
sec2 x + sec x tan x
=
.
sec x + tan x
sec x + tan x
This may seem pointless, but it will actually allow us to use a basic substitution to evaluate
the integral. Setting u = sec x + tan x so that du = sec x tan x + sec2 x, we have
Z
Z
sec2 x + sec x tan x
sec xdx =
dx
sec x + tan x
Z
1
=
du
u
= ln |u| + C
= ln | sec x + tan x| + C.
6