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MATH 118, LECTURE 3:
Trigonometric Integrals
1
Trigonometric Integrals
Many trigonometric integrals can be solved with simple manipulation.
We will break the problems for this lecture into three sections:
• integrals involving sin(x) and cos(x),
• integrals involving tan(x) and sec(x), and
• integrals involving cot(x) and csc(x).
The reason for this grouping is that differentiating the functions in these
three classes still gives functions within the same grouping (e.g. differentiating sin(x) or cos(x) gives functions consisting of sin(x) and cos(x); differentiating tan(x) and sec(x) gives functions consisting of tan(x) and sec(x);
etc.). In this lecture, we will make use of this fact to help us integrate
functions containing elements within these groupings.
1.1
Integrals with sin(x) and cos(x)
Consider the derivative of powers of sin(x) and cos(x), e.g.
d
sinn+1 (x) = (n + 1) sinn (x) cos(x).
dx
We can rearrange the constants and integrate to get
Z
1
sinn (x) cos(x) dx =
sinn+1 (x) + C, n 6= −1.
n+1
Similarly, for powers of cos(x) we have
Z
1
cosn+1 (x) + C, n 6= −1.
cosn (x) sin(x) dx = −
n+1
(1)
(2)
These integrals could also be derived by integration by substitution, choosing
u = sin(x) and u = cos(x) respectively (try it!).
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In other words, if we can manipulate integrals containing sin(x) and
cos(x) so that there is a product involving a single sin(x) or cos(x) term,
we can integrate the expression. We can handle the case when n = −1 by
substitution to complete (1) and (2):
Z
Z
−1
sin (x) cos(x) dx = cot(x) dx = ln | sin(x)| + C,
Z
cos
−1
(x) sin(x) dx =
Z
tan(x) dx = ln | sec(x)| + C.
In order to manipulate the equations into an integrable form, we will
need to make use of several identities:
sin2 (x) + cos2 (x) = 1,
(3)
sin(2x) = 2 sin(x) cos(x)
2
(4)
2
cos(2x) = 2 cos (x) − 1 = 1 − 2 sin (x).
(5)
Example 1:
Evaluate
Z
sin5 (x) dx.
We can see immediately that this does not satisfy the form of either (1) or
(2). We need to manipulate this, using our trigonometric identities, to get
an expression where there is only one term of cos(x) or sin(x) in a product.
We can do this by factoring out two terms of sin2 (x) and using identity (3)
to get
Z
Z
sin5 (x) dx =
sin(x)(1 − cos2 (x))(1 − cos2 (x)) dx
Z
=
sin(x) − 2 cos2 (x) sin(x) + cos4 (x) sin(x) dx
= − cos(x) +
1
2
cos3 (x) − cos5 (x) + C.
3
5
(You have to be careful with the signs here. Since the derivative of cos(x)
is − sin(x), the sign flips as a result of the chain rule.)
Example 2:
2
Evaluate
Z
cos3 (x)
dx.
sin1/2 (x)
The root (and fraction) may be a distraction here, but it does not limit our
ability to evaluate the integral since we need only to get one of the cos(x)
or sin(x) terms to the first power. It is clear from our identities, it cannot
be sin(x), so we have
Z
Z
cos3 (x)
dx =
sin−1/2 (x)(1 − sin2 (x)) cos(x) dx
sin1/2 (x)
Z
=
sin−1/2 (x) cos(x) − sin3/2 (x) cos(x) dx
= 2 sin1/2 (x) −
Example 3:
Evaluate
Z
2
sin5/2 (x) + C.
5
sin2 (x) cos2 (x) dx.
We might be first tempted to using identity (3); however, it will quickly
become obvious that no use of this identity will lead to a first power term of
either cos(x) or sin(x). Instead, we have to use identities (4) and (5). Using
a rearranged form of (5), we have
Z
Z
1
(1 + cos(2x))(1 − cos(2x)) dx
sin2 (x) cos2 (x) dx =
4
Z
1
=
(1 − cos2 (2x)) dx
4
Z 1
1 + cos(4x)
(Identity (5) again) =
1−
dx
4
2
Z
1
(1 − cos(4x)) dx
=
8
1
1
=
x−
sin(4x) + C.
8
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The text solves this problem using identity (4) (page 502, Trim).
1.2
Integrals with tan(x) and sec(x)
We can find similar properties using powers of tan(x) and sec(x). We have
d
tanm+1 (x) = (m + 1) tanm (x) sec2 (x), and
dx
3
d
secm (x) = m secm−1 (x) sec(x) tan(x) = m secm (x) tan(x).
dx
Rearranging the constants and integration, we arrive at the rules
Z
1
tanm+1 (x) + C, m 6= −1,
(6)
tanm (x) sec2 (x) dx =
m+1
Z
1
secm (x) tan(x) dx =
secm (x) + C, m 6= 0.
(7)
m
In other words, if we are able to manipulate equations of tan(x) and sec(x)
such that we have tan(x) to the first power or sec(x) to the second power,
we can immediately integrate. Note that the exceptional cases above (m 6=
−1, m 6= 0, respectively) simplify to the same cases as for sin(x) and cos(x)
(check!).
In order to manipulate the integrals, we have the identity
1 + tan2 (x) = sec2 (x).
Example 4:
Evaluate
Z
(8)
tan3 (x) sec3 (x) dx.
Recognizing that identity (8) shifts powers of tan(x) and sec(x) by two, we
need to factor out a tan2 (x).
Z
Z
3
3
tan (x) sec (x) dx =
(sec2 (x) − 1) sec3 (x) tan(x) dx
Z
=
sec5 (x) tan(x) − sec3 (x) tan(x) dx
=
Example 5:
Evaluate
Z
1
1
sec5 (x) − sec3 (x) + C.
5
3
sec4 (x)
dx.
tan2 (x)
It is best not to touch the demoninator, so we use identity (8) to get
Z
Z
sec4 (x)
dx =
tan−2 (x)(1 + tan2 (x)) sec2 (x) dx
tan2 (x)
Z
=
tan−2 (x) sec2 (x) + sec2 (x) dx
= − tan−1 (x) + tan(x) + C.
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1.3
Integrals with cot(x) and csc(x)
We can similarly derive the rules
Z
cotm csc2 (x) dx = −
Z
1
cotm+1 (x) + C, m 6= −1,
m+1
cscm (x) cot(x) dx = −
1
cscm (x) + C, m 6= 0.
m
(9)
(10)
Examples using cot(x) and csc(x) are very similar to those using tan(x)
and sec(x) and use the identity
1 + cot2 (x) = csc2 (x).
(11)
Example 6:
Evaluate
Z
3 + 4 csc2 (x)
dx.
cot2 (x)
This example is a little bit trickier, and we will also see that sometimes we
have to consider changing how we represent the functions (e.g. expanding
cot(x) into sin(x) and cos(x), etc.).
We will start be dividing out the denominator.
Z
Z
csc2 (x)
1
dx
+
4
dx.
3
cot2 (x)
cot2 (x)
The second integral is exactly in form of (9), so we can evaluate it directly
(being careful with the signs - there is one negative from the formula, and
one from the power of cot(x)) to get
Z
1
1
dx + 4
+ C.
3
cot2 (x)
cot(x)
It should be immediately clear that the first integral cannot be rearranged into a form satisfying (9) or (10). Our next line of reasoning is to
exchange cot(x) for sin(x)’s and cos(x)’s and see if we can manipulate the
equations then. This leads to
Z
sin2 (x)
1
3
dx + 4
+ C.
2
cos (x)
cot(x)
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This has not given us a form we can integrate (yet!), but we still have a few
things in out bag of tricks. In this case, we want to apply identity (3) on
the numerator to give us
Z
3 + 4 csc2 (x)
dx
cot2 (x)
Z
1 − cos2 (x)
1
=3
dx + 4
+C
2
cos (x)
cot(x)
Z
Z
1
dx − 3 1 dx + 4 tan(x) + C
=3
cos2 (x)
Z
= 3 sec2 (x) dx + 3x + 4 tan(x) + C
= 3 tan(x) + 3x + 4 tan(x) + C
= 7 tan(x) + 3x + C.
(NOTE: To be completely rigorous, I should have added a different constant
after each integrate and then pooled them at the end. This isn’t really
important - the important thing is the constant appears, at the very least,
after the final integration.)
It occurs to me now (after the lecture!) that we could have more easily
evaluated the troublesome integral by using tan(x) and sec(x) identities as
follows:
Z
Z
1
dx =
tan2 (x) dx
cot2 (x)
Z
=
[sec2 (x) − 1] dx
= tan(x) − x + C.
This just goes to show that there is often more than one way to solve a
problem. If you keep the identities available to us in mind, eventually the
solution should follow - one way or another.
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