LECTURE 17: TRIGONOMETRIC INTEGRALS
MINGFENG ZHAO
February 13, 2015
Theorem 1.
Z
 p+1

 x
+ C,
p+1
=

 ln |x| + C,
xp dx
Z
cos(ax) dx
=
sin(ax) dx
=
sec2 (ax) dx
=
Z
csc2 (ax) dx
Z
eax dx
=
Z
=
1
dx
a2 − x2
Z
1
dx
x2 + a2
√
if p = −1.
1
sin(ax) + C
a
1
− cos(ax) + C
a
1
tan(ax) + C
a
1
− cot(ax) + C
a
1 ax
e +C
a
x
sin−1
+C
a
x
1
tan−1
+C
a
a
Z
Z
if p 6= −1
=
=
Integration by parts
Z
Example 1. For any non-negative integer n, let In =
xn ex dx.
By the ‘ILATE’ rule, let
u = xn
dv = ex dx
=⇒ du = nxn−1 dx
Z
=⇒ v = ex dx = ex
Then
Z
In
=
xn ex dx
1
2
MINGFENG ZHAO
Z
=
u dv
Z
= uv − v du
Z
n x
= x e − nxn−1 ex
= xn ex − nIn−1 .
Z
Notice that I0 =
ex dx = ex , then
Z
I1 =
x
x
Z
x
xe dx = xe − e ,
I2 =
x2 ex dx = x2 ex − 2(xex − ex ) = x2 ex − 2xex + 2ex , · · · .
Trigonometric integrals
How to evaluate the following integrals:
Z
sinm (x) cosn (x) dx,
Z
and
tanm (x) secn (x) dx.
Here are some reduction formulas for mn = 0 in the above integrals: Assume n is a positive integer:
Z
Z
sinn−1 (x) cos(x) n − 1
+
sinn−2 (x) dx
sinn (x) dx = −
n
n
Z
Z
cosn−1 (x) sin(x) n − 1
cosn (x) dx =
+
cosn−2 (x) dx
n
n
Z
Z
tann−1 x
n
tan (x) dx =
− tann−2 (x) dx, n 6= 1
n−1
Z
Z
secn−2 (x) tan(x) n − 2
+
secn−2 (x) dx, n 6= 1.
secn (x) dx =
n−1
n−1
Remark 1. By the above reduction formulas, we only need to know the following integrals:
Z
Z
Z
Z
1 dx = x + C,
sin(x) dx = − cos(x) + C,
cos(x) dx = sin(x) + C, and
tan(x) dx = − ln | cos(x)| + C.
Z
Example 2. Evaluate
sec(x) dx.
Notice that
sec(x) =
1
cos(x)
1
cos(x)
cos(x)
=
·
=
=
.
cos(x)
cos(x) cos(x)
cos2 (x)
1 − sin2 (x)
So
Z
Z
sec(x) dx
=
cos(x)
dx
1 − sin2 (x)
LECTURE 17: TRIGONOMETRIC INTEGRALS
Z
=
=
1
du Let u = sin(x), then du = cos(x)dx
1 − u2
Z
1
1
1
−
+
du
2
u−1 u+1
1
(− ln |u − 1| + ln |u + 1|) + C
2
1 u + 1 +C
ln
=
2 u − 1
1 sin(x) + 1 =
+ C Since u = sin(x)
ln 2
sin(x) − 1 1 (1 + sin(x))2 =
+C
ln
2 1 − sin2 (x) 1 (1 + sin(x))2 =
+C
ln
2 cos2 (x) 1 + sin(x) +C
= ln cos(x) =
=
ln | sec(x) + tan(x)| + C.
Remark 2. During the computations of Example 2, we use the identity:
1
1
1
1
=
+
−
,
1 − u2
2
u−1 u+1
which will be studied in the section about the partial fractions, which is a very important topic.
Z
Example 3. Evaluate
csc(x) dx.
Notice that
csc(x) =
π
1
1
=
=
sec
−
x
.
sin(x)
2
cos π2 − x
Then
Z
Z
csc(x) dx
=
=
=
=
=
π
sec
− x dx
2
Z
π
− sec(u) du Let u = − x
2
− ln | sec(u) + tan(u)| + C By the result of Example 2
π
π
π
− ln sec
− x + tan
− x + C Since u = − x
2
2
2
− ln | csc(x) + cot(x)| + C.
3
4
MINGFENG ZHAO
Z
sinm (x) cosn (x) dx
Evaluate
Recall the half-angle formula:
sin2 (x) =
Z
To evaluate
1 − cos(2x)
,
2
and
cos2 (x) =
1 + cos(2x)
.
2
sinm (x) cosn (x) dx:
Cases
Strategy
m odd and positive, n any real number
Split off sin(x), rewrite the resulting even powers of sin(x) in terms of cos(x),
and then use u = cos(x)
n odd and positive, m any real number
Split off cos(x), rewrite the resulting even powers of cos(x) in terms of sin(x),
and then use u = sin(x)
m and n both even
Use half-angle formulas to transform the integrand into a polynomial in cos(2x),
and apply the preceding strategies once again to powers of cos(2x) greater than 1.
Z
Example 4. Evaluate
sin3 (x) cos−2 (x) dx.
This is the Case 1, we have
Z
3
−2
sin (x) cos
(x) dx
Z
sin2 (x) · sin(x) · cos−2 (x) dx
Z
sin2 (x) · cos−2 (x) · sin(x) dx
Z
(1 − cos2 (x)) cos−2 (x) · sin(x) dx
Z
(cos−2 (x) − 1) sin(x) dx
=
=
=
=
Z
=
=
=
=
=
(u−2 − 1) · (−1) du Let u = cos(x), then du = − sin(x)dx
Z
Z
−2
− u + 1 du
1
−u+C
u
1
− cos(x) + C
cos(x)
sec(x) − cos(x) + C.
Since u = cos(x)
LECTURE 17: TRIGONOMETRIC INTEGRALS
Z
Example 5. Evaluate
sin4 (x) cos2 (x) dx.
This is the Case 3, we have
Z
Z sin4 (x) cos2 (x) dx
=
1
8
=
=
=
Z
2 1 + cos(2x)
·
dx
2
Z
(1 − cos(2x) − cos2 (2x) + cos3 (2x)) dx
Z Z
1
1 + cos(4x)
1
(1 − sin2 (2x)) cos(2x) dx
1 − cos(2x) −
dx +
8
2
8
Z
1
1
1
1
1
1
sin2 (2x) cos(2x) dx
x−
sin(2x) − x −
sin(4x) +
sin(2x) −
8
16
16
64
16
8
Z
1
1
1
sin2 (2x) cos(2x) dx
x−
sin(4x) −
16
64
8
=
For
1 − cos(2x)
2
sin2 (2x) cos(2x) dx, use u = sin(2x), we get
Z
2
Z
sin (2x) cos(2x) dx =
u2
u3
1
du =
+ C = sin3 (2x) + C.
2
6
6
So we get
Z
Z
Evaluate
sin4 (x) cos2 (x) dx =
1
1
1
x−
sin(4x) −
sin3 (2x) + C.
16
64
48
tanm (x) secn (x) dx
Recall the trigonometric identities:
1 + tan2 (x) = sec2 (x),
Z
To evaluate
and
1 + cot2 (x) = csc2 (x).
tanm (x) secn (x) dx:
Cases
Strategy
n even
Split off sec2 (x), rewrite the remaining even power of sec(x) in terms of tan(x),
and use u = tan(x)
m odd
Split off sec(x) tan(x), rewrite the remaining even power of tan(x) in terms of sec(x),
and use u = sec(x)
m even and n odd
Rewrite the even power of tan(x) in terms of sec(x) to produce a polynomial in sec(x),
apply reduction formula to each term
5
6
MINGFENG ZHAO
Z
Example 6. Evaluate
tan3 (x) sec4 (x) dx.
This is both Case 1 and Case 2. Here we can use strategy 1, so we get
Z
Z
tan3 (x) sec4 (x) dx =
tan3 (x) · sec2 (x) · sec2 (x) dx
Z
=
tan3 (x)[1 + tan2 (x)] · sec2 (x) dx
Z
=
u3 (1 + u2 ) du Let u = tan(x), then du = sec2 (x)dx
Z
Z
=
u3 du + u5 du
=
=
1 4 1 6
u + u +C
4
6
1
1
tan4 (x) + tan6 (x) + C
4
6
Since u = tan(x).
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca