Trigonometric Integrals
This section investigates integrals of the form
Z
m
Z
n
sin (x) cos (x) dx
and
tanm (x) secn (x) dx.
In order to evaluate these integrals you MUST know the half-angle and Pythagorean identities:
sin2 (x) = 12 (1 − cos(2x))
1 + tan2 (x) = sec2 (x).
cos2 (x) = 21 (1 + cos(2x))
cos2 (x) + sin2 (x) = 1
Z
Example 1. Evaluate
π
sin2 (x) dx.
0
Intuition: In order to integrate sin2 (x) we will use the half-angle formula sin2 (x) =
express sin2 (x) in terms of two easier to integrate functions.
Z
π
2
sin (x) dx =
0
=
=
=
Z
1 π
1 − cos(2x) dx
2 0
π
1
1
x − sin(2x) 2
2
0
1
[(π − 0) − (0 − 0)]
2
π
2
See solution video
1
1
2
(1 − cos(2x)) to
Calculus II Resources
Integration Techniques
Guidelines for evaluating
R
sinm (x) cosn (x) dx
(1) If the power of cosine is odd (n = 2k + 1), save one cosine factor and use cos2 (x) = 1 − sin2 (x) to
express the remaining factors in terms of sine:
Z
Z
sinm (x) cos2k+1 (x) dx = sinm (x)(cos2 (x))k cos(x) dx
Z
= sinm (x)(1 − sin2 (x))k cos(x) dx.
Then substitute u = sin(x).
(2) If the power of sine is odd (m = 2k + 1), save one sine factor and use sin2 (x) = 1 − cos2 (x) to
express the remaining factors in terms of cosine:
Z
Z
sin2k+1 (x) cosn (x) dx = (sin2 (x))k sin(x) cosn (x) dx
Z
= (1 − cos2 (x))k cosn (x) sin(x) dx.
Then substitute u = cos(x).
(3) If the powers of both sine and cosine are even, use the identities
sin2 (x) =
Z
Example 2. Evaluate
1
(1 − cos(2x))
2
and
cos2 (x) =
1
(1 + cos(2x)) .
2
sin3 (x) cos2 (x) dx.
Intuition: Since the power
R of sine is odd we use the power reduction technique described in part (1) of the
Guidelines for evaluating sinm (x) cosn (x) dx.
Z
3
Z
2
sin (x) cos (x) dx =
Z
=
sin2 (x) cos2 (x) sin(x) dx
1 − cos2 (x) cos2 (x) sin(x) dx
We now use the substitution u = cos(x), so that du = − sin(x) and −du = sin(x).
Z
3
Z
2
sin (x) cos (x) dx =
1 − cos2 (x) cos2 (x) sin(x) dx
Z
=−
1 − u2 u2 du
Z
= u4 − u2 du
1 5 1 3
u − u +C
5
3
1
1
5
= cos (x) − cos3 (x) + C
5
3
Observation: This power reduction technique can be applied to ANY odd power of sine or cosine.
=
See solution video
2
Calculus II Resources
Integration Techniques
Z
Example 3. Evaluate
sec(x) dx.
Observation:
The integrand sec(x) does not fit any of the forms provided by the Guidelines for evaluating
R
tanm (x) secn (x) dx (See below). Evaluating this integral requires a bit of ingenuity.
R
In order to evaluate sec(x) dx we will multiply the integrand by 1 in a clever way and then use the
substitution u = sec(x) + tan(x) so that du = (sec(x) tan(x) + sec2 (x))dx.
Z
sec(x) + tan(x)
dx
sec(x) + tan(x)
Z
sec(x) tan(x) + sec2 (x)
=
dx u = sec(x) + tan(x)
sec(x) + tan(x)
Z
1
=
du
u
= ln |u| + C
Z
sec(x) dx =
sec(x)
= ln |sec(x) + tan(x)| + C
Observation: When encountering
R
sec(x) dx on future problems, do not recreate this work. The formula
Z
sec(x) dx = ln |sec(x) + tan(x)| + C.
is a common antiderivative that we must know.
See solution video
3
Calculus II Resources
Integration Techniques
Guidelines for evaluating
R
tanm (x) secn (x) dx
(1) If the power of secant is even (n = 2k),
Z
Z
tanm (x) sec2k (x) dx = tanm (x)(sec2 (x))k−1 sec2 (x) dx
Z
= tanm (x)(1 + tan2 (x))k−1 sec2 (x) dx.
Then substitute u = tan(x).
(2) If the power of tangent is odd (m = 2k + 1),
Z
Z
tan2k+1 (x) secn (x) dx = tan2k (x) secn−1 (x) sec(x) tan(x) dx
Z
= (sec2 (x) − 1)k secn−1 (x) sec(x) tan(x) dx.
Then substitute u = sec(x).
(3) If the power of tangent is even (m = 2k) and there is no secant term,
Z
Z
tan2k (x) dx = tan2(k−1) (x) tan2 (x) dx
Z
= tan2(k−1) (x)(sec2 (x) − 1) dx.
Algebraically expand and repeat if necessary.
(4) If there are no tangent terms and the power of secant is odd and larger than 1, apply IBPs.
(5) Try converting to sines and cosines.
Z
Example 4. Evaluate
tan3 (x) sec5 (x) dx.
Observation: Since the power of
R tangent is odd we use the power reduction technique described in part (2)
of the Guidelines for evaluating tanm (x) secn (x) dx.
Z
3
Z
5
tan (x) sec (x) dx =
Z
=
Z
=
Z
=
tan2 (x) sec4 (x) sec(x) tan(x) dx
(sec2 (x) − 1) sec4 (x) sec(x) tan(x) dx
(u2 − 1)u4 du
u6 − u4 du
1 7 1 5
u − u +C
7
5
1
1
7
= sec (x) − sec5 (x) + C
7
5
=
See solution video
4
u = sec(x)
du = sec(x) tan(x)dx
Calculus II Resources
Integration Techniques
See Trigonometric Integrals overview video
Practice Problems: Evaluate the following Integrals
Z
(1)
Z
(2)
Z
(3)
5
Z
cos (x) dx
(4)
tan(x) sec4 (x) dx
(5)
cos3 (x)
p
dx
sin(x)
(6)
Z
Z
See Solutions
5
sin2 (x) cos2 (x) dx
tan4 (x) dx
sec3 (x) dx (Hint: Use IBPs)