Section 7.2: Trigonemetric Integrals
1. Basic Trigonometric Integrals and Identities
In this section, we approach the problem of evaluating trigonometric
integrals (integrals involving powers and sums of the basic trigonometric functions). Before we do this, we recall a few facts that will be
extremely useful.
(i ) The basic trigonometric integrals we need:
(a)
Z
sin (x)dx = − cos (x) + C
(b)
Z
(c)
(d)
Z
Z
cos (x)dx = sin (x) + C
tan (x)dx = ln | sec (x)| + C
sec (x)dx = ln | sec (x) + tan (x)| + C
(ii ) The basic trig identities we need.
(a)
sin2 (x) + cos2 (x) = 1
(b)
sec2 (x) = 1 + tan2 (x)
(c)
1
cos2 (x) = (1 + cos (2x))
2
(d)
1
sin2 (x) = (1 − cos (2x))
2
(e)
sin (x) cos (x) =
1
sin (2x)
2
(f)
1
sin (A) cos (B) = [sin (A − B) + sin (A + B)]
2
(g)
1
sin (A) sin (B) = [cos (A − B) − cos (A + B)]
2
1
2
(h)
1
cos (A) cos (B) = [cos (A − B) + cos (A + B)]
2
It is not imperative that you memorize all of these formulas since there
are a lot, and many are very similar so it could cause confusion. On
a test, you will be able to use a sheet which includes these (and a few
other) formulas. What is important is learning the techniques involved
for a trigonometric integral. There are many circumstances which are
very clear cut, which we shall discuss in detail. However, there are
even more cases which are not quite so clear cut and require some nonstandard reasoning. The basic idea behind any trigonometric integral
is the following:
(i ) Try to simplify the integrand using trigonometric identities to
obtain an integrand on which you can apply integration by
substitution. This may require a number of steps or may fall
into one of a few special cases.
(ii ) Complete the integral using integration by substitution.
As you can probably imagine, a bulk of the work comes from the first
step of simplification, and it is what we shall spend most time on. We
shall proceed through cases.
2. Evaluating
R
sinm (x) cosn (x)dx
This falls into different cases depending upon whether the powers of
cos (x) and sin (x) are even or odd.
(i ) When the power of cos (x) is odd, we substitute cos2 (x) =
1 − sin2 (x) to eliminate all but a single power of cos (x) and
then substitute u = sin (x).
Example 2.1.
Z
Z
3
4
cos (x) sin (x)dx = cos(x)(1 − sin2 (x))(sin2 (x))dx
=
Z
cos(x)(sin2 (x) − sin4 (x))dx
Letting u = sin(x), we get du/dx = cos(x) or dx = du/ cos(x),
so
Z
Z
du
2
4
= cos(x)(sin (x) − sin (x))dx = cos(x)(u2 − u4 )
cos(x)
=
Z
(u2 − u4 )du =
u3 u5
sin3 (x) sin5 (x)
−
+C =
−
+C
3
5
3
5
3
(ii ) When the power of sin (x) is odd and the power of cos(x) is
even, we substitute sin2 (x) = 1 − cos2 (x) to eliminate all but
a single power of sin (x) and then substitute u = cos (x). It
works in exactly the same way as the previous case, so we shall
not go over another example.
(iii ) When both power of cos (x) and the power of sin(x) is even, we
use half angle identities repeatedly to reduce the trigonometric
equation into something we can integrate.
Example 2.2.
Z
Z
1
2
2
sin (x) cos (x)dx =
(1 − cos(2x))(1 + cos(2x))dx
4
Z
Z
1
1
1
2
1 + cos (2x)dx =
(1 + (1 + cos(4x)))dx
=
4
4
2
Z
1
3x sin(4x)
=
3 + cos(4x)dx =
+
+C
8
8
32
3. Evaluating
R
tanm (x) secn (x)dx
This falls into different cases depending upon whether the powers of
sec (x) and tan (x) are even or odd. For the case which do not fall
into one of the two categories below, there is no single clear cut way to
evaluate the given integral.
(i ) When the power of secant is even, save a factor of sec2 (x) and
use sec2 (x) = 1 + tan2 (x) to write all other factors in terms of
tan(x). Then substitute u = tan(x).
Z
Example 3.1.
4
3
sec (x) tan (x)dx =
=
Z
Z
sec2 (x)(1 + tan2 (x)) tan3 (x)dx
sec2 (x)(tan3 (x) + tan5 (x))dx
Substituting u = tan(x), we get dx = secdu
2 (x) , and
Z
Z
u4 u6
2
3
5
sec (x)(tan (x) + tan (x))dx = (u3 + u5 )du =
+
+C
4
6
=
tan4 (x) tan6 (x)
+
+C
4
6
(ii ) If the power of tangent is odd, save a factor of sec(x) tan(x)
and use tan2 (x) = sec2 (x)−1 to write all other factors in terms
of sec(x). Then substitute u = sec(x).
4
Example 3.2.
Z
Z
3
tan (x)dx = tan(x)(sec2 (x) − 1)dx
Z
Z
= sec(x) tan(x) sec(x)dx − tan(x)dx
The second integral can be evaluated explicitly. The first
integral requires the substitution u = sec(x), in which case
dx = sec(x)dutan(x) (WHY?). Thus,
Z
Z
Z
sec(x) tan(x) sec(x)dx − tan(x)dx = udu − ln | sec(x)|
u2
sec2 (x)
− ln | sec(x)| + C =
− ln | sec(x)| + C
2
2
R
4. Evaluating cos(mx) sin(nx)dx and other such integrals
=
For these types of integrals, we use the sum and difference formulas.
Since they all follow the same idea, we illustrate with just one example.
Example 4.1.
Z
Z
1
sin(3x) cos(4x)dx =
(sin(3x − 4x) + sin(3x + 4x))dx
2
Z
1
cos(7x)
1
(− sin(x) + sin(7x))dx = (cos(x) −
)+C
=
2
2
7
5. Other Trigonometric Integrals
Most other trigonometric functions are not so rigid, and often completely different techniques may be required for other types. We finish
by looking at a non standard example.
Example 5.1.
Z
Z
sin2 (x)
cos (x) tan (x)dx = cos2 (x) 2 dx
cos (x)
Z
Z
x sin(2x)
1
(1 − cos(2x))dx = −
+C
= sin2 (x)dx =
2
2
4
2
2