Trigonometric Integrals Solutions
Practice Problems: Evaluate the following Integrals
Z
(1)
Z
(2)
Z
(3)
Z
cos5 (x) dx
(4)
tan(x) sec4 (x) dx
(5)
cos3 (x)
p
dx
sin(x)
Z
Z
(6)
sin2 (x) cos2 (x) dx
tan4 (x) dx
sec3 (x) dx (Hint: Use IBPs)
Solutions
Z
(1) Evaluate
cos5 (x) dx
Observation: Since the power of
R cosine is odd we use the power reduction technique described in part (2)
of the Guidelines for evaluating sinm (x) cosn (x) dx.
Z
Z
5
cos (x) dx =
Z
=
Z
=
Z
=
Z
=
cos4 (x) cos(x) dx
(cos2 (x))2 cos(x) dx
(1 − sin2 (x))2 cos(x) dx
u = sin(x)
(1 − u2 )2 du
u4 − 2u2 + 1 du
1 5 2 3
u − u +u+C
5
3
1
2
5
= sin (x) − sin3 (x) + sin(x) + C
5
3
=
See solution video
Z
(2) Evaluate
tan(x) sec4 (x) dx
Observation: Since the power of
R secant is even we use the power reduction technique described in part (1)
of the Guidelines for evaluating tanm (x) secn (x) dx.
1
Calculus II Resources
Integration Techniques
Z
Z
4
tan(x) sec (x) dx =
Z
=
Z
=
tan(x) sec2 (x) sec2 (x) dx
tan(x)(1 + tan2 (x)) sec2 (x) dx
u = tan(x)
u + u3 du
1 2 1 4
u + u +C
2
4
1
1
2
= tan (x) + tan4 (x) + C
2
4
=
See solution video
Z
(3) Evaluate
cos3 (x)
p
dx
sin(x)
Observation: Based upon the form of the integrand we would like to try a u-Substitution. In order to
choose u as an “inside” function we set u = sin(x) so that du = cos(x)dx. To convert into an integral ONLY
involving the variable u we apply the power reduction technique to cos3 (x).
Z
cos3 (x)
p
dx =
sin(x)
Z
cos2 (x)
p
cos(x) dx
sin(x)
Z
1 − sin2 (x)
p
cos(x) dx
sin(x)
Z
1 − u2
√
du
u
Z
u−1/2 − u3/2 du
=
=
=
2
= 2u1/2 − u5/2 + C
5
p
2
= 2 sin(x) − sin5/2 (x) + C.
5
See solution video
Z
(4) Evaluate
sin2 (x) cos2 (x) dx
Observation: Since the powers of sine and cosine are both even we rewrite the integrand with the half-angle
formulas
cos2 (2) =
1
(1 + cos(22))
2
and
2
sin2 (2) =
1
(1 − cos(22)) .
2
Calculus II Resources
Integration Techniques
Z
2
2
sin (x) cos (x) dx =
=
=
=
=
=
Z
1
(1 − cos(2x)(1 + cos(2x)) dx
4
Z
1
1 − cos2 (2x) dx
4
Z
Z
1
1
1 dx −
cos2 (2x) dx
4
4
Z
1
1
1
x−
(1 + cos(4x)) dx
4
4
2
1
1
1
x − [x + sin(4x)] + C
4
8
4
1
1
x−
sin(4x) + C.
8
32
See solution video
Z
(5) Evaluate
tan4 (x) dx
Observation: Since the power of tangent is even and
R no secant term is present we apply the technique
described in part (3) of the Guidelines for evaluating tanm (x) secn (x) dx.
Z
tan4 (x) dx =
Z
tan2 (x) tan2 (x) dx
Z
tan2 (x)(sec2 (x) − 1) dx
Z
Z
= tan2 (x) sec2 (x) dx − tan2 (x) dx
Z
Z
2
= u du − sec2 (x) − 1 dx
=
u = tan(x)
1 3
u − tan(x) + x + C
3
1
= tan3 (x) − tan(x) + x + C
3
=
See solution video
Z
(6) Evaluate
sec3 (x) dx
Observation: Since the tangent term is not present and the power of secant is odd we apply IBPs. Expressing sec3 (x) = sec(x) sec2 (x) and observing that we know an antiderivative of sec2 (x) suggests we choose
u = sec(x) and dv = sec2 (x).
Using
u = sec(x)
du = sec(x) tan(x)dx
−→
.
v = tan(x)
IBPs gives
dv = sec2 (x)dx
3
Calculus II Resources
Z
Integration Techniques
sec3 (x) dx = sec(x) tan(x) −
Z
sec(x) tan2 (x) dx
Z
sec(x)(sec2 (x) − 1) dx
Z
Z
= sec(x) tan(x) − sec3 (x) dx + sec(x) dx
Z
= sec(x) tan(x) − sec3 (x) dx + ln | sec(x) + tan(x)|.
= sec(x) tan(x) −
Adding
R
sec3 (x) dx to both sides and dividing by 2 yields
Z
sec(x) tan(x) + ln | sec(x) + tan(x)|
sec3 (x) dx =
+ C.
2
See solution video
4