Trigonometric Integrals
In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine.
EXAMPLE 1 Evaluate
3
y cos x dx.
SOLUTION Simply substituting u cos x isn’t helpful, since then du sin x dx. In order
to integrate powers of cosine, we would need an extra sin x factor. Similarly, a power of
sine would require an extra cos x factor. Thus, here we can separate one cosine factor
and convert the remaining cos2x factor to an expression involving sine using the identity
sin 2x cos 2x 1:
cos 3x cos 2x cos x 1 sin 2x cos x
We can then evaluate the integral by substituting u sin x, so du cos x dx and
y cos x dx y cos x cos x dx y 1 sin x cos x dx
3
2
2
y 1 u 2 du u 13 u 3 C
sin x 13 sin 3x C
In general, we try to write an integrand involving powers of sine and cosine in a form
where we have only one sine factor (and the remainder of the expression in terms of
cosine) or only one cosine factor (and the remainder of the expression in terms of sine).
The identity sin 2x cos 2x 1 enables us to convert back and forth between even powers
of sine and cosine.
EXAMPLE 2 Find
5
2
y sin x cos x dx
SOLUTION We could convert cos 2x to 1 sin 2x, but we would be left with an expression in
terms of sin x with no extra cos x factor. Instead, we separate a single sine factor and
rewrite the remaining sin 4x factor in terms of cos x :
sin 5x cos 2x sin2x2 cos 2x sin x 1 cos 2x2 cos 2x sin x
Figure 1 shows the graphs of the integrand
sin 5x cos 2x in Example 2 and its indefinite integral (with C 0). Which is which?
■ ■
Substituting u cos x, we have du sin x dx and so
y sin x cos x dx y sin x
5
2
2
2
cos 2x sin x dx
0.2
y 1 cos 2x2 cos 2x sin x dx
_π
π
_0.2
FIGURE 1
y 1 u 2 2 u 2 du y u 2 2u 4 u 6 du
u3
u5
u7
2
3
5
7
C
13 cos 3x 25 cos 5x 17 cos 7x C
1
2 ■ TRIGONOMETRIC INTEGRALS
In the preceding examples, an odd power of sine or cosine enabled us to separate a
single factor and convert the remaining even power. If the integrand contains even powers
of both sine and cosine, this strategy fails. In this case, we can take advantage of the following half-angle identities (see Equations 17b and 17a in Appendix C):
sin 2x 12 1 cos 2x
EXAMPLE 3 Evaluate
y
0
cos 2x 12 1 cos 2x
and
sin 2x dx.
SOLUTION If we write sin 2x 1 cos 2x, the integral is no simpler to evaluate. Using the
half-angle formula for sin 2x, however, we have
y
0
[ (x sin 2x dx 12 y 1 cos 2x dx 1
2
0
1
2
0
]
sin 2x)
12 ( 12 sin 2) 12 (0 12 sin 0) 12 Notice that we mentally made the substitution u 2x when integrating cos 2x. Another
method for evaluating this integral was given in Exercise 33 in Section 5.6.
1.5
y=sin@ x
Example 3 shows that the area of the
region shown in Figure 2 is 2.
■ ■
π
0
FIGURE 2
_0.5
EXAMPLE 4 Find
4
y sin x dx.
SOLUTION We could evaluate this integral using the reduction formula for
x sin n x dx
(Equation 5.6.7) together with Example 3 (as in Exercise 33 in Section 5.6), but a better
method is to write sin 4x sin 2x2 and use a half-angle formula:
y sin x dx y sin x dx
4
2
y
2
1 cos 2x
2
2
dx
14 y 1 2 cos 2x cos 2 2x dx
Since cos 2 2x occurs, we must use another half-angle formula
cos 2 2x 12 1 cos 4x
This gives
y sin x dx y 1 2 cos 2x 4
1
4
14 y
1
2
1 cos 4x dx
( 32 2 cos 2x 12 cos 4x) dx
4 ( 2 x sin 2x 8 sin 4x) C
1 3
1
To summarize, we list guidelines to follow when evaluating integrals of the form
x sin mx cos nx dx, where m 0 and n 0 are integers.
TRIGONOMETRIC INTEGRALS ■ 3
Strategy for Evaluating
y sin
m
x cos nx dx
(a) If the power of cosine is odd n 2k 1, save one cosine factor and use
cos 2x 1 sin 2x to express the remaining factors in terms of sine:
y sin
m
x cos 2k1x dx y sin m x cos 2xk cos x dx
y sin m x 1 sin 2xk cos x dx
Then substitute u sin x.
(b) If the power of sine is odd m 2k 1, save one sine factor and use
sin 2x 1 cos 2x to express the remaining factors in terms of cosine:
y sin
x cos n x dx y sin 2xk cos n x sin x dx
2k1
y 1 cos 2xk cos n x sin x dx
Then substitute u cos x. [Note that if the powers of both sine and cosine are
odd, either (a) or (b) can be used.]
(c) If the powers of both sine and cosine are even, use the half-angle identities
sin 2x 12 1 cos 2x
cos 2x 12 1 cos 2x
It is sometimes helpful to use the identity
sin x cos x 12 sin 2x
We can use a similar strategy to evaluate integrals of the form x tan mx sec nx dx. Since
ddx tan x sec 2x, we can separate a sec 2x factor and convert the remaining (even)
power of secant to an expression involving tangent using the identity sec 2x 1 tan 2x.
Or, since ddx sec x sec x tan x, we can separate a sec x tan x factor and convert the
remaining (even) power of tangent to secant.
EXAMPLE 5 Evaluate
6
4
y tan x sec x dx.
SOLUTION If we separate one sec 2x factor, we can express the remaining sec 2x factor in
terms of tangent using the identity sec 2x 1 tan 2x. We can then evaluate the integral
by substituting u tan x with du sec 2x dx :
y tan x sec x dx y tan x sec x sec x dx
6
4
6
2
2
y tan 6x 1 tan 2x sec 2x dx
y u 61 u 2 du y u 6 u 8 du
u7
u9
C
7
9
17 tan 7x 19 tan 9x C
4 ■ TRIGONOMETRIC INTEGRALS
EXAMPLE 6 Find
y tan sec d.
5
7
SOLUTION If we separate a sec 2 factor, as in the preceding example, we are left with
a sec 5 factor, which isn’t easily converted to tangent. However, if we separate a
sec tan factor, we can convert the remaining power of tangent to an expression
involving only secant using the identity tan 2 sec 2 1. We can then evaluate the
integral by substituting u sec , so du sec tan d :
y tan 5
sec 7 d y tan 4 sec 6 sec tan d
y sec 2 12 sec 6 sec tan d
y u 2 12 u 6 du y u 10 2u 8 u 6 du
u 11
u9
u7
2
C
11
9
7
111 sec 11 29 sec 9 17 sec 7 C
The preceding examples demonstrate strategies for evaluating integrals of the form
x tan mx sec nx dx for two cases, which we summarize here.
Strategy for Evaluating
y tan
m
x sec nx dx
(a) If the power of secant is even n 2k, k 2, save a factor of sec 2x and use
sec 2x 1 tan 2x to express the remaining factors in terms of tan x :
y tan
m
x sec 2kx dx y tan m x sec 2xk1 sec 2x dx
y tan m x 1 tan 2xk1 sec 2x dx
Then substitute u tan x.
(b) If the power of tangent is odd m 2k 1, save a factor of sec x tan x and
use tan 2x sec 2x 1 to express the remaining factors in terms of sec x :
y tan
x sec n x dx y tan 2xk sec n1x sec x tan x dx
2k1
y sec 2x 1k sec n1x sec x tan x dx
Then substitute u sec x.
For other cases, the guidelines are not as clear-cut. We may need to use identities, integration by parts, and occasionally a little ingenuity. We will sometimes need to be able to
integrate tan x by using the formula established in Example 5 in Section 5.5:
y tan x dx ln sec x C
TRIGONOMETRIC INTEGRALS ■ 5
We will also need the indefinite integral of secant:
y sec x dx ln sec x tan x C
1
We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x tan x :
sec x tan x
y sec x dx y sec x sec x tan x dx
y
sec 2x sec x tan x
dx
sec x tan x
If we substitute u sec x tan x, then du sec x tan x sec 2x dx, so the integral
becomes x 1u du ln u C. Thus, we have
y sec x dx ln sec x tan x C
EXAMPLE 7 Find
3
y tan x dx.
SOLUTION Here only tan x occurs, so we use tan 2x sec 2x 1 to rewrite a tan 2x factor in
terms of sec 2x :
y tan x dx y tan x tan x dx
3
2
y tan x sec 2x 1 dx
y tan x sec 2x dx y tan x dx
tan 2x
ln sec x C
2
In the first integral we mentally substituted u tan x so that du sec 2x dx.
If an even power of tangent appears with an odd power of secant, it is helpful to express
the integrand completely in terms of sec x. Powers of sec x may require integration by
parts, as shown in the following example.
EXAMPLE 8 Find
3
y sec x dx.
SOLUTION Here we integrate by parts with
u sec x
du sec x tan x dx
Then
dv sec 2x dx
v tan x
y sec x dx sec x tan x y sec x tan x dx
3
2
sec x tan x y sec x sec 2x 1 dx
sec x tan x y sec 3x dx y sec x dx
6 ■ TRIGONOMETRIC INTEGRALS
Using Formula 1 and solving for the required integral, we get
y sec x dx (sec x tan x ln sec x tan x ) C
1
2
3
Integrals such as the one in the preceding example may seem very special but they
occur frequently in applications of integration, as we will see in Chapter 6. Integrals of
the form x cot m x csc n x dx can be found by similar methods because of the identity
1 cot 2x csc 2x.
Finally, we can make use of another set of trigonometric identities:
2 To evaluate the integrals (a) x sin mx cos nx dx, (b) x sin mx sin nx dx, or
(c) x cos mx cos nx dx, use the corresponding identity:
(a) sin A cos B 12 sinA B sinA B
These product identities are discussed in
Appendix C.
■ ■
(b) sin A sin B 12 cosA B cosA B
(c) cos A cos B 12 cosA B cosA B
EXAMPLE 9 Evaluate
y sin 4x cos 5x dx.
SOLUTION This integral could be evaluated using integration by parts, but it’s easier to use
the identity in Equation 2(a) as follows:
y sin 4x cos 5x dx y
1
2
sinx sin 9x dx
12 y sin x sin 9x dx
12 (cos x 19 cos 9x C
Exercises
A Click here for answers.
1–47
Evaluate the integral.
3
2
1.
y sin x cos x dx
3.
y
34
2
sin 5x cos 3x dx
5
4
5.
y cos x sin x dx
7.
y
9.
11.
13.
Click here for solutions.
S
2
0
y
0
cos2 d
sin 43t dt
2
y
4
0
4.
y
d
sin 4x cos 2x dx
2
0
cos 5x dx
6.
y sin mx dx
8.
y
12.
14.
2
y
0
sin 2 2 d
cos6 d
2
y x cos x dx
y
2
0
sin 2x cos 2x dx
18.
y cot sin d
20.
y cos x sin 2x dx
22.
y
2
24.
y tan x dx
6
26.
y
tan 5 x sec 4 x dx
28.
y tan 2x sec 2x dx
3
30.
y
32.
y tan ay dy
17.
y cos x tan x dx
19.
y
21.
y sec x tan x dx
23.
y tan x dx
25.
y sec t dt
27.
y
29.
y tan x sec x dx
31.
y tan x dx
2
3
1 sin x
dx
cos x
2
3
0
y cos cos sin d
y sin x scos x dx
3
y sin x cos x dx
10.
y 1 cos 6
2.
16.
3
15.
3
0
5
5
5
4
2
2
0
sec 4t2 dt
4
4
0
sec 4 tan 4 d
3
3
0
5
tan 5x sec6x dx
6
TRIGONOMETRIC INTEGRALS ■ 7
tan 3
33.
y cos d
35.
y
4
2
6
cot 2x dx
; 57–58
Use a graph of the integrand to guess the value of the
integral. Then use the methods of this section to prove that your
guess is correct.
2
34.
y tan x sec x dx
36.
y
2
4
cot 3x dx
57.
y
2
0
■
37.
y cot 3 csc 3 d
38.
y csc 4 x cot 6 x dx
39.
y csc x dx
40.
y
41.
3
6
42.
y sin 5x sin 2x dx
43.
y cos 7 cos 5 d
45.
y
1 tan x
dx
sec 2x
47.
y t sec t
y
cos x sin x
dx
sin 2x
2
■
■
2
■
■
■
■
■
■
62. y cos x, y 0, x 0, x 2;
about y 1
51.
■
■
■
■
■
■
y sec
■
4
■
■
■
■
■
■
■
■
■
■
■
65–67
■
■
■
53. Find the average value of the function f x sin 2x cos 3x on
the interval , .
54. Evaluate x sin x cos x dx by four methods: (a) the substitution
Prove the formula, where m and n are positive integers.
y sin mx cos nx dx 0
66.
y sin mx sin nx dx 67.
y cos mx cos nx dx ■
■
■
■
0
if m n
if m n
if m n
if m n
0
■
■
■
■
■
■
68. A finite Fourier series is given by the sum
N
f x a
n
sin nx
n1
y sin 3x,
56. y sin x,
y 2 sin x,
■
a 1 sin x a 2 sin 2x a N sin Nx
Find the area of the region bounded by the given curves.
55. y sin x,
■
■
x 0,
2
■
■
■
x 2
■
Show that the mth coefficient a m is given by the formula
x 2
x 0,
■
■
am ■
■
65.
■
u cos x, (b) the substitution u sin x, (c) the identity
sin 2x 2 sin x cos x, and (d) integration by parts. Explain the
different appearances of the answers.
55–56
■
where t is the time in seconds. Voltmeters read the RMS (rootmean-square) voltage, which is the square root of the average
value of Et 2 over one cycle.
(a) Calculate the RMS voltage of household current.
(b) Many electric stoves require an RMS voltage of 220 V.
Find the corresponding amplitude A needed for the voltage
Et A sin120 t.
x
dx
2
■
■
current that varies from 155 V to 155 V with a frequency
of 60 cycles per second (Hz). The voltage is thus given by
the equation
Et 155 sin120 t
■
4
■
■
64. Household electricity is supplied in the form of alternating
y sin x cos x dx
52.
y sin 3x sin 6x dx
■
4
■
f 0 0.
Evaluate the indefinite integral. Illustrate, and check that
your answer is reasonable, by graphing both the integrand and its
antiderivative (taking C 0.
50.
■
about y 1
; 49–52
5
■
61. y cos x, y 0, x 0, x 2;
x04 tan 8 x sec x dx in terms of I.
y sin x dx
■
about the x-axis
48. If x04 tan 6 x sec x dx I , express the value of
49.
■
63. A particle moves on a straight line with velocity function
vt sin t cos 2 t. Find its position function s f t if
y cos x 1
■
■
60. y tan 2x, y 0, x 0, x 4;
2
■
■
sin 2 x cos 5 x dx
about the x-axis
tan t dt
4
■
2
0
59. y sin x, x 2, x , y 0;
■
dx
46.
■
y
Find the volume obtained by rotating the region bounded
by the given curves about the specified axis.
csc 3x dx
2
■
58.
59–62
y sin 3x cos x dx
44.
■
cos 3x dx
■
■
■
1
y
f x sin mx dx
■
■
8 ■ TRIGONOMETRIC INTEGRALS
Answers
49. 5 cos 5x 3 cos 3x cos x C
1
S
Click here for solutions.
2
1.1
ƒ
1.
5.
11.
15.
17.
21.
25.
29.
31.
33.
37.
41.
45.
1
5
1
5
11
cos 5x 13 cos 3x C
3. 384
2
1
5
7
9
sin x 7 sin x 9 sin x C
7. 4
9. 38
3
1
13.
2
sin
sin
2
C
3
4192
2
4
( 27 cos 3x 23 cos x) scos x C
1
2
19. ln1 sin x C
2 cos x ln cos x C
1
2
23. tan x x C
2 tan x C
1
2
117
5
3
27. 8
5 tan t 3 tan t tan t C
1
3
3 sec x sec x C
1
4
2
4 sec x tan x ln sec x C
1
1
6
4
35. s3 3
6 tan 4 tan C
1
1
3
5
39. ln csc x cot x C
csc
csc
C
3
5
1
1
1
1
43. 4 sin 2 24 sin 12 C
6 sin 3x 14 sin 7x C
1
1
5 2
47. 10 tan t C
2 sin 2x C
F
_2π
1.1
51.
1
6
sin 3x 181 sin 9x C
1
ƒ
F
2π
_2
2
1
1
3
55.
57. 0
53. 0
63. s 1 cos 3 t3
59. 24
61. 2 24
TRIGONOMETRIC INTEGRALS ■ 9
Solutions: Trigonometric Integrals
s
c
The symbols = and = indicate the use of the substitutions {u = sin x, du = cos x dx} and {u = cos x, du = − sin x dx},
respectively.
¢
R¡
¢
c R¡
sin2 x cos2 x sin x dx = 1 − cos2 x cos2 x sin x dx = 1 − u2 u2 (−du)
¢
R¡
¢
R¡
= u2 − 1 u2 du = u4 − u2 du = 15 u5 − 13 u3 + C = 15 cos5 x − 13 cos3 x + C
R 3π/4
R 3π/4
R 3π/4
¡
¢
3. π/2 sin5 x cos3 x dx = π/2 sin5 x cos2 x cos x dx = π/2 sin5 x 1 − sin2 x cos x dx
√
¢
R √2/2 ¡ 5
¢
£
¤√2/2
s R 2/2 5 ¡
u − u7 du = 16 u6 − 18 u8 1
= 1
u 1 − u2 du = 1
´ ¡
³
¢
11
− 16 − 18 = − 384
= 1/8
− 1/16
6
8
1.
5.
R
R
sin3 x cos2 x dx =
¢2
R¡
¢2
s R ¡
cos4 x sin4 x cos x dx =
1 − sin2 x sin4 x cos x dx =
1 − u2 u4 du
¢
¢
R¡
R
¡
=
1 − 2u2 + u4 u4 du =
u4 − 2u6 + u8 du = 15 u5 − 27 u7 + 19 u9 + C
R
cos5 x sin4 x dx =
R
=
7.
9.
11.
13.
15.
17.
R π/2
0
Rπ
0
R
cos2 θ dθ =
=
1
5
sin5 x −
R π/2
0
£
1
2
2
7
sin7 x +
1
9
sin9 x + C
1
(1
2
θ+
1
2
0
R
(1 + cos θ)2 dθ = (1 + 2 cos θ + cos2 θ) dθ = θ + 2 sin θ +
= θ + 2 sin θ +
R π/4
0
R
Z
π
4
Rπ
¤2
Rπ£
¤2
sin2 (3t) dt = 0 12 (1 − cos 6t) dt = 14 0 (1 − 2 cos 6t + cos2 6t) dt
Rπ£
Rπ¡
¤
¢
= 14 0 1 − 2 cos 6t + 12 (1 + cos 12t) dt = 14 0 32 − 2 cos 6t + 12 cos 12t dt
£
£¡
¤π
¢
¤
1
= 14 32 t − 13 sin 6t + 24
sin 12t 0 = 14 3π
− 0 + 0 − (0 − 0 + 0) = 3π
2
8
sin4 (3t) dt =
Rπ£
+ cos 2θ) dθ [half-angle identity]
£¡
¤π/2
¢
¤
sin 2θ 0 = 12 π2 + 0 − (0 + 0) =
sin3 x
2
+
1
4
sin 2θ + C =
3
2θ
R
(1 + cos 2θ) dθ
+ 2 sin θ +
0
sin2 x (sin x cos x)2 dx =
R π/4
1
4
sin 2θ + C
¡
¢2
− cos 2x) 12 sin 2x dx
R π/4
R π/4
R π/4
= 18 0 (1 − cos 2x) sin2 2x dx = 18 0 sin2 2x dx − 18 0 sin2 2x cos 2x dx
R π/4
£
£
¤π/4
¤π/4
1
1 1
1
x − 14 sin 4x − 13 sin3 2x 0
(1 − cos 4x) dx − 16
sin3 2x 0 = 16
= 16
3
0
¡
¢
1 π
1
1
= 192
= 16
(3π − 4)
4 −0− 3
sin4 x cos2 x dx =
R π/4
1
2θ
1
2
0
1
2 (1
´
¢
R¡
¢√
R ³ 5/2
√
c R ¡
cos x dx =
1 − cos2 x cos x sin x dx =
1 − u2 u1/2 (−du) =
u
− u1/2 du
3
= 27 u7/2 − 23 u3/2 + C = 27 (cos x)7/2 −
¢√
¡
= 27 cos3 x − 23 cos x cos x + C
cos x tan x dx =
Z
sin3 x
c
dx =
cos x
Z ¡
2
3
(cos x)3/2 + C
¢
¸
Z ·
1 − u2 (−du)
−1
=
+ u du
u
u
= − ln |u| + 12 u2 + C =
1
2
cos2 x − ln |cos x| + C
10 ■ TRIGONOMETRIC INTEGRALS
19.
Z
1 − sin x
dx =
cos x
Z
(sec x − tan x) dx = ln |sec x + tan x| − ln |sec x| + C
·
¸
by (1) and the boxed
formula above it
= ln |(sec x + tan x) cos x| + C = ln |1 + sin x| + C
Or:
Z
= ln (1 + sin x) + C since 1 + sin x ≥ 0
¢
Z
Z ¡
Z
1 − sin2 x dx
1 − sin x 1 + sin x
cos x dx
1 − sin x
dx =
·
dx =
=
cos x
cos x
1 + sin x
cos x (1 + sin x)
1 + sin x
Z
dw
=
[where w = 1 + sin x, dw = cos x dx]
w
= ln |w| + C = ln |1 + sin x| + C = ln (1 + sin x) + C
R
sec2 x tan x dx = u du = 12 u2 + C = 12 tan2 x + C.
R
R
Or: Let v = sec x, dv = sec x tan x dx. Then sec2 x tan x dx = v dv = 12 v 2 + C = 12 sec2 x + C.
21. Let u = tan x, du = sec2 x dx. Then
23.
25.
27.
R
tan2 x dx =
R
sec6 t dt =
R
R¡
R
¢
sec2 x − 1 dx = tan x − x + C
R
R
sec4 t · sec2 t dt = (tan2 t + 1)2 sec2 t dt = (u2 + 1)2 du
R
= (u4 + 2u2 + 1) du = 15 u5 + 23 u3 + u + C =
R π/3
0
tan5 x sec4 x dx =
=
=
1
5
tan5 t +
2
3
[u = tan t, du = sec2 t dt]
tan3 t + tan t + C
R π/3
tan5 x (tan2 x + 1) sec2 x dx
0
√
R 3 5 2
u (u + 1) du
[u = tan x, du = sec2 x dx]
0
√
¤√3
£
R 3 7
27
81
(u + u5 ) du = 18 u8 + 16 u6 0 = 81
8 + 6 = 8
0
+
9
2
=
81
8
+
36
8
=
117
8
Alternate solution:
R π/3
R π/3
R π/3
tan5 x sec4 x dx = 0 tan4 x sec3 x sec x tan x dx = 0 (sec2 x − 1)2 sec3 x sec x tan x dx
0
R2
[u = sec x, du = sec x tan x dx]
= 1 (u2 − 1)2 u3 du
R2
R2 4
2
3
= 1 (u − 2u + 1)u du = 1 (u7 − 2u5 + u3 ) du
¤2 ¡
¢
£
¢ ¡
= 18 u8 − 13 u6 + 14 u4 1 = 32 − 64
+ 4 − 18 − 13 + 14 = 117
3
8
29.
R
tan3 x sec x dx =
1 3
u
3
−u+C =
R¡
¢
sec2 x − 1 sec x tan x dx
[u = sec x, du = sec x tan x dx]
1
3
sec3 x − sec x + C
¢2
R
R
R
sec2 x − 1 tan x dx = sec4 x tan x dx − 2 sec2 x tan x dx + tan x dx
R
R
R
= sec3 x sec x tan x dx − 2 tan x sec2 x dx + tan x dx
R
tan5 x dx =
=
33.
tan2 x sec x tan x dx =
R
= (u2 − 1) du
=
31.
R
Z
R¡
1
4
tan3 θ
dθ =
cos4 θ
=
sec4 x − tan2 x + ln |sec x| + C
Z
R
R
tan3 θ sec4 θ dθ =
u3 (u2 + 1) du
Z
[or
37.
R π/2
π/6
R
cot2 x dx =
R π/2 ¡
π/6
cot3 α csc3 α dα =
R
sec4 x − sec2 x + ln |sec x| + C ]
tan3 θ · (tan2 θ + 1) · sec2 θ dθ
[u = tan θ, du = sec2 θ dθ]
= (u5 + u3 ) du = 16 u6 + 14 u4 + C =
35.
1
4
1
6
tan6 θ +
1
4
¢
¡
π/2
csc2 x − 1 dx = [− cot x − x]π/6 = 0 −
tan4 θ + C
π
2
¢
¡ √
− − 3−
π
6
¢
=
√
3−
R
cot2 α csc2 α · csc α cot α dα = (csc2 α − 1) csc2 α · csc α cot α dα
R
[u = csc α, du = − csc α cot α dα]
= (u2 − 1)u2 · (−du)
R 2
4
1 3
1 5
= (u − u ) du = 3 u − 5 u + C = 13 csc3 α − 15 csc5 α + C
π
3
TRIGONOMETRIC INTEGRALS ■ 11
Z
− csc x cot x + csc2 x
csc x (csc x − cot x)
dx =
dx. Let u = csc x − cot x ⇒
csc x − cot x
csc x − cot x
¢
R
¡
du = − csc x cot x + csc2 x dx. Then I = du/u = ln |u| = ln |csc x − cot x| + C.
39. I =
Z
csc x dx =
Z
41. Use Equation 2(b):
R
R
sin 5x sin 2x dx =
=
1
6
1
[cos(5x
2
sin 3x −
− 2x) − cos(5x + 2x)] dx =
1
14
sin 7x + C
1
2
R
(cos 3x − cos 7x) dx
43. Use Equation 2(c):
R
R
cos 7θ cos 5θ dθ =
45.
Z
R
1
1
(cos 2θ + cos 12θ) dθ
2 [cos(7θ − 5θ) + cos(7θ + 5θ)] dθ = 2
¡
¢
1 1
1
1
1
= 2 2 sin 2θ + 12 sin 12θ + C = 4 sin 2θ + 24
sin 12θ + C
1 − tan2 x
dx =
sec2 x
Z
¡
¢
cos2 x − sin2 x dx =
Z
cos 2x dx =
47. Let u = tan(t2 ) ⇒ du = 2t sec2 (t2 ) dt. Then
¡ ¢
¡ ¢
R
¡
¢
R
1 5
u +C =
t sec2 t2 tan4 t2 dt = u4 12 du = 10
1
10
1
sin 2x + C
2
tan5 (t2 ) + C.
49. Let u = cos x ⇒ du = − sin x dx. Then
¢2
R¡
¢2
R¡
R
1 − cos2 x sin x dx =
1 − u2 (−du)
sin5 x dx =
R¡
¢
=
−1 + 2u2 − u4 du = − 15 u5 + 23 u3 − u + C
= − 15 cos5 x +
2
3
cos3 x − cos x + C
Notice that F is increasing when f (x) > 0, so the graphs serve as a
check on our work.
51.
R
=
R
=
1
6
sin 3x sin 6x dx =
1
[cos(3x
2
R
1
2
− 6x) − cos(3x + 6x)] dx
(cos 3x − cos 9x) dx
sin 3x −
1
18
sin 9x + C
Notice that f (x) = 0 whenever F has a horizontal tangent.
53. fave =
1
2π
=
1
2π
=0
¡
¢
Rπ
1
sin2 x cos3 x dx = 2π
sin2 x 1 − sin2 x cos x dx
−π
R0 2¡
¢
[where u = sin x]
u 1 − u2 du
0
Rπ
−π
55. For 0 < x < π2 , we have 0 < sin x < 1, so sin3 x < sin x. Hence the area is
R π/2 ¡
¢
R π/2
¡
¢
R π/2
sin x − sin3 x dx = 0 sin x 1 − sin2 x dx = 0 cos2 x sin x dx. Now let u = cos x ⇒
0
R0
¤1
R1
£
du = − sin x dx. Then area = 1 u2 (−du) = 0 u2 du = 13 u3 0 = 13 .
R 2π
It seems from the graph that 0 cos3 x dx = 0, since the area below the
57.
x-axis and above the graph looks about equal to the area above the axis
and below the graph. By Example 1, the integral is
£
¤2π
sin x − 13 sin3 x 0 = 0. Note that due to symmetry, the integral of
any odd power of sin x or cos x between limits which differ by 2nπ
(n any integer) is 0.
12 ■ TRIGONOMETRIC INTEGRALS
£
¤π
¡
Rπ
π sin2 x dx = π π/2 12 (1 − cos 2x) dx = π 12 x − 14 sin 2x π/2 = π π2 − 0 −
¤
R π/2 ¡
R π/2 £
¢
(1 + cos x)2 − 12 dx = π 0
2 cos x + cos2 x dx
61. Volume = π 0
59. V =
Rπ
π/2
π
4
¢
¤π/2
¡
£
2
= π 2 sin x + 12 x + 14 sin 2x 0 = π 2 + π4 = 2π + π4
Rt
63. s = f (t) = 0 sin ωu cos2 ωu du. Let y = cos ωu ⇒ dy = −ω sin ωu du. Then
R cos ωt 2
£
¤cos ωt
¡
¢
1
1 − cos3 ωt .
y dy = − ω1 13 y3 1
= 3ω
s = − ω1 1
65. Just note that the integrand is odd [f (−x) = −f (x)].
Or: If m 6= n, calculate
Rπ
−π
sin mx cos nx dx =
=
Rπ
1
[sin(m
−π 2
− n)x + sin(m + n)x] dx
·
¸π
cos(m − n)x
cos(m + n)x
1
−
−
=0
2
m−n
m+n
−π
If m = n, then the first term in each set of brackets is zero.
Rπ
Rπ
67. −π cos mx cos nx dx = −π 21 [cos(m − n)x + cos(m + n)x] dx. If m 6= n,
·
¸π
1 sin(m − n)x
sin(m + n)x
this is equal to
= 0. If m = n, we get
+
2
m−n
m+n
−π
¸π
·
Rπ 1
£ 1 ¤π
sin(m + n)x
[1
+
cos(m
+
n)x]
dx
=
x
+
= π + 0 = π.
2
−π 2
−π
2(m + n) −π
¢
+0 =
π2
4