Trigonometric Integrals - Bruce E. Shapiro

Topic 7
Trigonometric Integrals
In this section we will tackle integrals primarily of the forms
∫
sin m x cos n x dx (7.1)
and
∫
sec m x tan n x dx (7.2)
where m and n are integrals, as well as integrals of the form
∫
∫
∫
sin px sin qx dx, sin px cos qx dx, cos px cos qx dx (7.3)
where p and q are integrals. Our primary tools will be the basic trigonometric
identities
cos 2 x + sin 2 x = 1 (7.4)
The double angle formula for cosines is
cos 2x = cos 2 x − sin 2 x (7.5)
Adding equations (7.5) and (7.4) and dividing by 2 gives
cos 2 x = 1 (1 + cos 2x) (7.6)
2
Similarly, subtracting equation (7.5) from (7.4) and dividing by 2 gives
sin 2 x = 1 (1 − cos 2x) (7.7)
2
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Math 150A TOPIC 7. TRIGONOMETRIC INTEGRALS
If we divide equation (7.4) by cos x we get a similar equation relating the
tangent and secant of an angle:
Finally, we add the double angle formula for a sine
tan 2 x + 1 = sec 2 x (7.8)
sin 2x = 2 sin x cos x (7.9)
and the following product-angle formulas to the mix:
sin x cos y = 1 [sin(x − y) + sin(x + y)] (7.10)
2
sin x sin y = 1 [cos(x − y) − cos(x + y)] (7.11)
2
cos x cos y = 1 [cos(x − y) + cos(x + y)] (7.12)
2
Equations (7.4) through (7.12) will be our main tools for finding substitutions
that reduce integrals to manageable proportions.
∫
Example 7.1 sin 2 x dx This is perhaps the most basic of the integrals
of this form. We can use the identity (7.7) to obtain
∫
sin 2 x dx = 1 ∫
(1 − cos 2x) dx (7.13a)
2
= 1 2 x − 1 sin 2x + C (7.13b)
4
When there is a single odd power (e.g., cos 3 x, sin 3 x, cos 5 x, sin 5 x, . . . ), we
“reserve” one factor for the derivative, and replace all the other factors with
the identity (7.4), followed by a u-substitution of u = cos x or u = sin x.
∫
Example 7.2 cos 3 x dx
∫
∫
cos 3 x dx = cos x · cos 2 x dx
(7.14a)
∫
= cos x (1 − sin 2 x) dx
(7.14b)
∫
= (1 − u 2 ) du where u = sin x (7.14c)
= u − 1 3 u3 + C (7.14d)
= sin x − 1 3 sin3 x + C (7.14e)
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TOPIC 7. TRIGONOMETRIC INTEGRALS Math 150A
∫
Example 7.3
sin 5 x dx
∫
∫
sin 5 x dx = sin x(sin 2 x) 2 dx
(7.15a)
∫
= sin x(1 − cos 2 x) 2 dx
(7.15b)
= −
∫
(1 − u 2 ) 2 du with u = cos x (7.15c)
= −
∫
(1 − 2u 2 + u 4 )du (7.15d)
= −u + 2 3 u3 − 1 5 u5 + C (7.15e)
= − cos x + 2 3 cos3 x − 1 5 cos5 x + C (7.15f)
When there are purely even powers, the substitutions (7.6) and (7.7) reduce
the power of the exponent. In the following example, we have to use the
substitution twice. The first time, we substitute sin 2 x = (1/2)(1−cos(2x)).
The second time, we use cos 2 (2x) = (1/2)(1 + cos(4x))
∫
Example 7.4 sin 4 x dx
∫
∫
sin 4 x dx =
(sin 2 x) 2 dx
(7.16a)
∫ ( ) 2 1
=
2 (1 − cos 2x) dx (7.16b)
= 1 ∫
(1 − 2 cos(2x) + cos 2 (2x)) dx (7.16c)
4
= 1 4 dx − 1 ∫
cos(2x) dx + 1 ∫
cos 2 (2x)dx (7.16d)
2
4
= x 4 − 1 4 sin 2x + 1 ∫
(1 + cos(4x))dx (7.16e)
8
= x 4 − 1 4 sin 2x + x 8 + 1 sin(4x) + C
32
(7.16f)
= 3x 8 − 1 4 sin 2x + 1 sin(4x) + C
32
(7.16g)
When there are a combination of sines and cosines we often have to use a
mix of these rules.
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Math 150A TOPIC 7. TRIGONOMETRIC INTEGRALS
∫
Example 7.5
sin 5 x cos 2 x dx
∫
∫
sin 5 x cos 2 x dx = sin x(sin 2 x) 2 cos 2 x
(7.17a)
∫
= sin x(1 − cos 2 x) 2 cos 2 x dx
(7.17b)
= −
∫
(1 − u 2 ) 2 u 2 du with u = cos x (7.17c)
= −
∫
(u 2 − 2u 4 + u 6 ) du (7.17d)
= − 1 3 u3 + 2 5 u5 − 1 7 u7 + C (7.17e)
= − 1 3 cos3 x + 2 5 cos5 x − 1 7 cos7 x + C (7.17f)
∫
Example 7.6
sin 4 x cos 2 x dx
∫
∫
sin 4 x cos 2 x dx =
⎛
⎞
sin 2 x
{ }} {
1
⎜
⎝2 (1 − cos(2x)) ⎟
⎠
2
( 1
2 (1 + cos(2x)) )
dx (7.18a)
∫ (1
− cos(2x) − cos 2 (2x) + cos 3 (2x) ) dx (7.18b)
= 1 8
= x 8 − 1
16 sin(2x) − 1 8
∫
cos 2 (2x) + 1 8
∫
cos 3 (2x) dx
(7.18c)
From exercise 7.1 we have
∫
cos 2 xdx = x 2 + 1 4 sin(2x)
(7.18d)
and from example 7.2
∫
cos 3 xdx = sin x − 1 3 sin3 x
(7.18e)
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TOPIC 7. TRIGONOMETRIC INTEGRALS Math 150A
hence
∫
sin 4 x cos 2 x dx = x 8 − 1
16 sin(2x) − 1 ( 2x
8 2 + 1 )
4 sin(4x)
+ 1 (
sin(2x) − 1 )
8
3 sin3 (2x)
(7.18f)
= 1 16 sin(2x) − 1
32 sin(4x) − 1 24 sin3 (2x) + C (7.18g)
∫
sin m x cos n x dx
1. If the exponent of the cosine is odd,
(a) Save one cosine for the derivative
(b) Replace all the other cosines with cos 2 x = 1 − sin 2 x
(c) Make a u-substitution of the form u = sin x
2. Otherwise, if the exponent of the sine is odd
(a) Save one sine for the derivative
(b) Replace all the other sines with sin 2 x = 1 − cos 2 x
(c) Make a u-substitution of the form u = cos x
3. Otherwise, if both the exponents of the sine and the cosine are even,
then use the following substitutions as possible and as necessary to
reduce the degree of the exponent:
sin 2 x = 1 (1 − cos 2x)
2
cos 2 x = 1 (1 + cos 2x)
2
sin 2x = 2 sin x cos x
The process for solving products of powers of secants and tangents is similar.
The idea is to use the identity
in combination with the fact that
tan 2 x + 1 = sec 2 x
d
dx tan x = sec2 x
d
sec x = sec x tan x
dx (7.19c)
(7.19a)
(7.19b)
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Math 150A TOPIC 7. TRIGONOMETRIC INTEGRALS
and an appropriate u-substitution of either u = sec x or u = tan x to
solve the integral. Unfortunately, the process is complicated by the fact
that, unlike in the case of sines and cosines, we do not have double-angle
formulas that reduce the powers. Thus our combination of substitutions
will not always work. It is helpful to recall the basic integration formulas
here as well:
∫
∫
tan x dx = ln | sec x| + C
sec x dx = ln | sec x + tan x| + C
(7.20a)
(7.20b)
∫
Example 7.7
tan 3 x dx
∫
∫
tan 3 x dx =
∫
=
∫
=
tan x tan 2 x dx
tan x(sec 2 x − 1) dx
∫
tan x sec 2 x dx − tan x dx
(7.21a)
(7.21b)
(7.21c)
= 1 2 tan2 xdx − ln | sec x| + C (7.21d)
where we have used u = tan x in the first integral.
∫
Example 7.8 tan 6 x sec 4 x dx
∫
∫
tan 6 x sec 4 x dx =
∫
=
∫
=
tan 6 x sec 2 x sec 2 x dx
tan 6 x(1 + tan 2 x) sec 2 x dx
∫
tan 6 x sec 2 x dx + tan 8 x sec 2 x dx
(7.22a)
(7.22b)
(7.22c)
= 1 7 tan7 x + 1 9 tan9 x + C (7.22d)
where we have used u = tan x in each integral.
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TOPIC 7. TRIGONOMETRIC INTEGRALS Math 150A
∫
Example 7.9
tan 5 x sec 7 x dx
∫
d(sec x)
∫
{ }} {
tan 5 x sec 7 x dx = tan 4 x sec 6 x (sec x tan x dx)
∫
= tan 4 x sec 6 x d(sec x)
∫
= (sec 2 x − 1) 2 sec 6 x d(sec x)
∫
= (sec 4 x − 2 sec 2 x + 1) sec 6 x d(sec x)
∫
= (sec 10 x − 2 sec 8 x + sec 6 x) ( .
sec x)
(7.23a)
(7.23b)
(7.23c)
(7.23d)
(7.23e)
= 1
11 sec11 x − 2 9 sec9 x + 1 7 sec7 x + C (7.23f)
Sometimes we can get somewhere by combining integration by parts with
an appropriate substitution, as in the following.
∫
Example 7.10 sec 3 x dx We start by using integration by parts, with
u = sec x =⇒ du = sec x tan x dx (7.24a)
dv = sec 2 x dx =⇒ v = tan x (7.24b)
With these substitutions
∫
∫
sec 3 x dx = uv − v du
∫
= sec x tan x − tan x sec x tan x dx
∫
= sec x tan x − tan 2 x sec x dx
∫
= sec x tan x − (sec 2 x − 1) sec x dx
∫
∫
= sec x tan x − sec 3 x dx + sec x dx
(7.24c)
(7.24d)
(7.24e)
(7.24f)
(7.24g)
Since the original integral appears on both sides of the equation, but with
opposite signs, we can add.
∫
∫
2 sec 3 x dx = sec x tan x + sec x dx
(7.24h)
= sec x tan x + ln | sec x + tan x| (7.24i)
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Math 150A TOPIC 7. TRIGONOMETRIC INTEGRALS
Hence
∫
sec 3 x dx = 1 (sec x tan x + ln | sec x + tan x|) + C
2 (7.24j)
Now we collect some rules for products of secants and tangents.
∫
tan m x sec n x dx
1. If the exponent on the tangent is even
(a) Save one factor of sec 2 x to use as d tan x.
(b) Substitute sec 2 x = 1 + tan 2 x in the remaining secants.
(c) Use a substitution u = tan x and integrate.
2. Otherwise, if the exponent of the secant is odd and the exponent of
the tangent is at least 1,
(a) Save one factor sec x tan x to use as d sec x.
(b) Substitute tan 2 x = sec 2 x − 1 for all the remaining tangents.
(c) Use a substitution u = sec x and integrate.
3. Otherwise, look for some other way to simplify the integral.
Finally, we also consider simple products of multiple-angle formulas, which
we demonstrate by example.
Example 7.11
∫ π/4
0
cos(2x) sin(12x) dx We use the trigonometric identity
sin θ cos ψ = 1 [sin(θ − ψ) + sin(θ + ψ)]
2 (7.25a)
to obtain
∫ π/4
0
∫ π/4
cos(2x) sin(12x) dx = 1 [sin(10x) + sin(14x)] dx (7.25b)
2 0
= 1 [ ]∣ −1
−1 ∣∣∣
π/4
cos(10x) +
2 10 14 cos(14x) (7.25c)
0
= 1 [ −1
2 10 cos 5π 2 − 1
14 cos 7π 2 + 1
10 + 1 ]
14
(7.25d)
= 3 35
(7.25e)
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TOPIC 7. TRIGONOMETRIC INTEGRALS Math 150A
Exercises
∫
7.1. cos 2 x dx
∫
7.2. cos 4 (3x) dx
∫
7.3. sin 3 (5x) dx
∫
7.4. cos 5 (x + 7) dx
∫ π/2
7.5. sin 2 (3x) dx
0
∫ π
7.6. cos 4 x dx
0
∫
7.7. sin 2 x cos 2 x dx
∫
7.8. sin x cos 3 x dx
∫
7.9. sin 4 x cos 5 x dx
∫
7.10. tan 2 (πx)dx
7.11.
7.12.
7.13.
7.14.
7.15.
7.16.
7.17.
7.18.
7.19.
7.20.
∫ 1
−1
tan 4 (πx/4)dx
∫
sec x tan 2 x dx
∫
sec 3 x tan x dx
∫ π/4
sec 3 x tan 3 x dx
0
∫
sec x tan 3 x dx
∫
sec 7 x tan 9 x dx
∫
sec 4 x tan 7 x dx
∫
sec 4 x tan 4 x dx
∫ π/2
∫
0
sin(3x) sin(6x) dx
sin(100x) cos(101x) dx
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Math 150A TOPIC 7. TRIGONOMETRIC INTEGRALS
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