6.2 Trigonometric Integrals and Substitutions
Arkansas Tech University MATH 2924: Calculus II
Dr. Marcel B. Finan
6.2 Trigonometric Integrals and Substitutions
In this section, we discuss integrals with trigonometric integrands and integrals that can be transformed to trigonometric integrals by substitution.
Trigonometric Integrals In order to understand the following discussion, the reader is encouraged to review trigonometric identities before proceeding with the discussion.
Integrals of the form sinn xdx or cosn xdx
Example 6.2.1 Find sin3 xdx.
Solution. First, we notice that sin3 x = sin x sin2 x = sin x(1 - cos2 x). Using the
substitution u = cos x, where du = - sin xdx, we find
sin3 xdx = - (1 - u2)du = -u + u3 + C = - cos x + cos3 x + C
3
3
Example 6.2.2 Find cos4 xdx.
Solution. Using the trigonometric identity
cos2 x = 1 + cos 2x 2
1
we find
cos4 xdx = [cos2 x]2dx =
1 + cos 2x 2 dx
2
1 =
(1 + 2 cos 2x + cos2 2x)dx
4
1 = dx + 2 cos 2xdx +
4
1 + cos 4x dx
2
1
sin 2x x 1 sin 4x
= x+2
++
4
2 2 24
13
sin 4x
= x + sin 2x +
42
8
Example 6.2.3 Find sin2 xdx.
Solution. Using the trigonometric identity
sin2 x = 1 - cos 2x 2
we find
sin2 xdx =
1 =
2 1 = 2
1 - cos 2x dx
2
(1 - cos 2x)dx
sin 2x
x-
+C
2
Integrals of the form sinn x cosm xdx
Example 6.2.4 Find sin3 x cos4 xdx.
2
Solution. Let u = cos x then du = - sin xdx. Thus,
sin3 x cos4 xdx = (1 - cos2 x) sin x cos4 xdx
= - (1 - u2)u4du = (-u4 + u6)du u5 u7
=- + +C 57
Example 6.2.5 Find sin2 x cos2 xdx.
cos5 x cos7 x
=-
+
+C
5
7
Solution. Using the trigonometric identity cos2 x + sin2 x = 1, we have
sin2 x cos2 xdx = sin2 x(1 - sin2 x)dx
= sin2 xdx - sin4 xdx
=
x sin 2x -
-
1 -
sin3
x
cos
x
+
3
sin2 xdx
24
4
4
x =
-
sin
2x
+
1
sin3
x
cos
x
-
3
x sin 2x -
+C
244
42 4
x =
-
sin
2x
+
1
sin3
x
cos
x
+
3
3 sin 2x - x + C
244
16
8
=-
1
sin
2x
+
1
sin3
x
cos
x
+
1 x
+
C
16
4
8
Integrals of Tangent and Secant
Example 6.2.6 Find tan xdx.
3
Solution. Using the substitution u = cos x, we find
sin x
du
tan xdx =
dx = -
cos x
u
= - ln |u| + C = - ln | cos x| + C
= ln | sec x| + C
Example 6.2.7 Find sec xdx.
Solution. First, note that
(sec x + tan x) = sec2 x + sec x tan x = sec x(sec x + tan x).
Using the substitution u = sec x + tan x, we find
sec xdx = =
sec x + tan x
sec x
dx
sec x + tan x
sec2 x + sec x tan x dx
sec x + tan x
du
=
= ln |u| + C
u
= ln | sec x + tan x| + C
4
Trigonometric Substitutions
Thissubsection deals with integrands involving terms like x2 - a2, x2 + a2,
and a2 - x2.
? Integrands involving a2 - x2, -a x a, a > 0.
For
each
x
in
the
interval
[-a,
a]
there
is
a
in
the
interval
[-
2
,
2
]
such
that
x=
a sin
(notice
that
-1
x a
1 and
recall
the
graph
of sin x).
Thus,
using
the
substitution
x
=
a
sin
,
-
2
2
to
obtain
a2 - x2 = a2(1 - sin2 )
= a2 cos2 = a| cos |
=a cos
where we have used the Pythagorean identity cos2 + sin2 = 1. Notice that
cos2
=
cos
since
cos
0
in
-
2
2
.
It
is
important
to
point
out
here that by constructing a right triangle with one of the angle being then
the and
hypotenuse of the the adjacent side
triangle haslength a, has length a2 - x2.
the opposite side has length x
It follows that cos =
a2-x2 a
.
See Figure 6.2.1.
Figure 6.2.1
Example 6.2.8
Find
1 4-x2
dx.
Solution.
Let
x
=
2
sin
,
-
2
<
<
2
.
Then
4 - x2 = 4 - 4 sin2 = 4 cos2 = 2 cos .
Moreover, dx = 2 cos d. It follows that
1
2 cos
x
dx =
d = + C = arcsin + C
4 - x2
2 cos
2
5
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