Download Integration of Trigonometric Functions

Integration of
Trigonometric
Functions
13.6
Introduction
Integrals involving trigonometric functions are commonplace in engineering mathematics. This
is especially true when modelling waves, and alternating current circuits. When the root-meansquare (rms) value of a waveform, or signal is to be calculated, you will often find this results
in an integral of the form
sin2 t dt
In this Section you will learn how such integrals can be evaluated.
'
$
① be able to find a number of simple definite
and indefinite integrals
Prerequisites
Before starting this Section you should . . .
&
Learning Outcomes
After completing this Section you should be
able to . . .
② be able to use a table of integrals
③ be familiar with standard trigonometric
identities
✓ use trigonometric identities to write integrands in alternative forms to enable
them to be integrated
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1. Integration of Trigonometric Functions
Simple integrals involving trigonometric functions have already been dealt with in Section 13.1.
See what you can remember:
Write
integrals:
down the following
a) sin x dx,
b) cos x dx,
c) sin 2x dx,
d)
cos 2x dx
Your solution
a) − cos x + c, b) sin x + c, c) − 21 cos 2x + c,
d)
1
2
sin 2x + c.
The basic rules from which these results can be derived are summarised here:
Key Point
cos kx
+c
sin kx dx = −
k
cos kx dx =
sin kx
+c
k
In engineering applications it is often necessary to integrate functions involving powers of the
trigonometric functions such as
2
or
cos2 ωt dt
sin x dx
Note that these integrals cannot be obtained directly from the formulas in the Key Point above.
However by making use of trigonometric identities the integrands can be re-written in an alternative form. It is often not clear which identities are useful and each case needs to be considered
individually. Experience and practice are essential. Work through the following Guided Exercise.
Use the trigonometric identity
to express the integral
sin2 θ = 12 (1 − cos 2θ)
sin2 x dx in an alternative form.
HELM (VERSION 1: March 18, 2004): Workbook Level 1
13.6: Integration of Trigonometric Functions
2
Your solution
The integral can be written
1
(1
2
− cos 2x)dx.
Note from the last exercise that the trigonometric identity was used to convert a power of sin x
into a function involving cos 2x which can be integrated directly using the Key Point above.
Now find the indefinite integral
sin2 x dx.
Your solution
x − 21 sin 2x + c = 21 x − 41 sin 2x + K where K = c/2.
1
2
Use the trigonometric identity sin 2x = 2 sin x cos x to find
sin x cos x dx
Your solution
The integrand can be written as
1
2
sin 2x whose indefinite integral is is − 41 cos 2x + c
Using the result of the previous example write down the value of
2π
sin x cos x dx
0
Your solution
3
HELM (VERSION 1: March 18, 2004): Workbook Level 1
13.6: Integration of Trigonometric Functions
2π
2π
1
sin x cos x dx = − cos 2x + c
4
0
0
1
1
= − cos 4π + cos 0
4
4
1 1
= − + =0
4 4
This result is one example of what are called orthogonality relations.
2. Orthogonality Relations
In general two functions f (x), g(x) are said to be orthogonal to each other over an interval
a ≤ x ≤ b if
b
f (x)g(x) dx = 0
a
It follows from the previous example that sin x and cos x are orthogonal to each other over the
interval 0 ≤ x ≤ 2π or indeed any other interval α ≤ x ≤ α + 2π (e.g. π/2 ≤ x ≤ 5π or
−π ≤ x ≤ π).
More generally there is a whole set of orthogonality relations involving these trigonometric
functions on intervals of length 2π (i.e. over one period of both sin x and cos x). These relations
are useful in connection with a widely used technique in engineering, known as Fourier analysis
where we represent periodic functions in terms of an infinite series of sines and cosines called a
Fourier series.
We shall demonstrate the orthogonality property
2π
Imn =
sin mx sin nx dx = 0
0
where m and n are integers such that m = n.
The secret is to use a trigonometric identity to convert the integrand into a form that can be
readily integrated.
You may recall the identity
1
sin A sin B = (cos(A − B) − cos(A + B))
2
It follows, putting A = mx and B = nx that
1 2π
Imn =
[cos(m − n)x − cos(m + n)x] dx
2 0
2π
1 sin(m − n)x sin(m + n)x
=
−
2
(m − n)
(m + n)
0
= 0
because (m − n) and (m + n) will be integers and sin(integer)×2π) = 0. Also of course sin 0 = 0.
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13.6: Integration of Trigonometric Functions
4
Why does the case m = n have to be excluded from the analysis?
The corresponding orthogonality relation for cosines
2π
cos mx cos nx dx = 0
Jmn =
0
follows by use of a similar identity to that just used. Here again m and n are integers such that
m = n.
Use the identity
1
sin A cos B = (sin(A + B) + sin(A − B))
2
to show that
2π
sin mx cos nx dx = 0
m and n integers, m = n.
Kmn =
0
Your solution
(recalling that cos((integer) × 2π) = 1)
= 0
We have, by the given identity,
1 2π
[sin(m + n)x + sin(m − n)x] dx
Kmn =
2 0
2π
1
cos(m + n)x cos(m − n)x
=
−
−
2
(m + n)
(m − n)
0
1 cos(m + n)2π − 1 cos(m − n)2π − 1
+
2
(m + n)
(m − n)
= −
Finally show that the orthogonality relation Kmn also holds if m = n. Hint:
You will need to use a different trigonometric identity.
5
HELM (VERSION 1: March 18, 2004): Workbook Level 1
13.6: Integration of Trigonometric Functions
Your solution
Note that the particular case m = n = 1 was considered earlier in this Section.
Putting m = n, and then using the identity sin 2A = 2 sin A cos A we get
2π
sin mx cos mx dx
Kmm =
0
1 2π
sin 2mx dx
=
2 0
2π
1
cos 2mx
=
−
2
2m
0
1
= −
(cos 4mπ − cos 0)
4m
1
(1 − 1) = 0
4m
= −
0
Kmn =
sin mx cos mx dx
2π
3. Reduction Formulae
You have seen earlier in this Workbook how to integrate sin x and sin2 x (which is sin x multiplied
by itself). Applications sometimes arise which involve integrating higher powers of sin x or cos x.
It is possible, as we now show, to obtain a reduction formula to aid in this task.
So consider
In = sinn (x) dx
Write down the precise integrals represented by I2 , I3 , I10
HELM (VERSION 1: March 18, 2004): Workbook Level 1
13.6: Integration of Trigonometric Functions
6
Your solution
I2 =
sin2 x dx
I3 =
sin3 x dx
I10 =
sin10 x dx
To obtain a reduction formula for In we write
sinn x = sinn−1 (x) sin x
and use integration by parts.
In the notation used earlier
in this Workbook put f = sinn−1 x and g = sin x
df
and evaluate
and g dx.
dx
Your solution
g dx =
sin x dx = − cos x
We have
df
= (n − 1) sinn−2 x cos x (using the chain rule of differentiation)
dx
Now use the integration by parts formula on
sinn−1 x sin x dx. (See earlier in the Workbook for
the parts formula if necessary). Do not attempt to evaluate the second integral that you obtain.
Your solution
= sinn−1 (x)(− cos x) + (n − 1)
g dx −
sinn−1 x sin x dx = sinn−1 (x)
df
dx
sinn−2 x cos2 x dx
g dx
7
HELM (VERSION 1: March 18, 2004): Workbook Level 1
13.6: Integration of Trigonometric Functions
Putting cos2 x = 1 − sin2 x in the integral on the right-hand side, this integral becomes:
n−2
sin (x) dx − sinn (x) dx
so finally
In =
sin
n−1
(x) sin x dx = sin
n−1
(x)(− cos x)+(n−1)
sin
n−2
(x) dx−(n−1)
sinn (x) dx
or
In = − sinn−1 (x) cos x + (n − 1)In−2 − (n − 1)In
from which
1
n−1
(∗)
In−2
In = − sinn−1 (x) cos x +
n
n
This is our reduction formula for In . It enables us, for example, to evaluate I6 in terms of I4 ,
then I4 in terms of I2 and indeed I2 in terms of I0 where
0
I0 = sin x dx = 1 dx = x.
Use the reduction formula with n = 2 to calculate I2 .
Your solution
(NOTE: here and elsewhere we are omitting the constant of integration.)
as obtained earlier by a different technique.
i.e.
1
1
I2 = − [sin x cos x] + I0
2
2
1 1
x
= − [ sin 2x] +
2 2
2
1
x
sin2 x dx = − sin 2x +
4
2
Use the reduction formula (∗) to obtain I6 =
sin6 x dx. Firstly obtain I6 in
terms of I4 , then I4 in terms of I2 . (You evaluated I2 in the previous exercise.)
HELM (VERSION 1: March 18, 2004): Workbook Level 1
13.6: Integration of Trigonometric Functions
8
Your solution
1
3
I4 = − sin3 x cos x + I2
4
4
1
5
I6 = − sin5 x cos x + I4
6
6
Then, by (∗) with n = 4
Using (∗) with n = 6
Now substitute for I2 from the previous exercise to obtain I4 and hence I6 .
Your solution
∴
5
1
5
5
I6 = − sin5 x cos x −
sin3 x cos x −
sin 2x + x
6
24
32
16
I4 = − 41 sin3 x cos x −
3
16
sin 2x + 83 x
Definite integrals can also be readily evaluated using the reduction formula (∗). For example,
π/2
sin nx dx
In =
0
π/2
sinn−2 x dx
so In−2 =
0
We obtain, immediately
π/2 n − 1
1
− sinn−1 (x) cos x 0 +
In−2
In =
n
n
or, since cos π2 = sin 0 = 0,
(n − 1)
In−2
n
This simple easy-to-use formula is well known and is called Wallis’ formula.
In =
9
HELM (VERSION 1: March 18, 2004): Workbook Level 1
13.6: Integration of Trigonometric Functions
π/2
sinn x dx calculate I1 and then use Wallis’ formula, without
If In =
0
further integration, to obtain I3 and I5 .
Your solution
π/2
8
4
4 2
sin5 x dx = I3 = × =
5
5 3
15
0
π/2
2
2
2
sin3 x dx = I1 = × 1 =
3
3
3
0
I5 =
I3 =
Then using Wallis’ formula with n = 3 and n = 5 respectively
I1 =
π/2
0
sin x dx = [− cos x]0
π/2
=1
The total power P of an antenna is given by
π
ηL2 I 2 π
sin3 θ dθ
P =
2
4λ
0
where η, λ, Iare constants as is the length L of antenna. Using the reduction
formula for
sinn x , obtain P .
Your solution
HELM (VERSION 1: March 18, 2004): Workbook Level 1
13.6: Integration of Trigonometric Functions
10
0
If I1 =
I3 =
π
ηL2 I 2 π
4λ2
0
then by the reduction formula (∗) with n = 3
π 2
4
1
2
− sin2 x cos x 0 + I1 = × 2 =
I3 =
3
3
3
3
π
ηL2 I 2 π
3λ2
Hence P =
sin3 θ dθ =
sin θ dθ = [− cos θ]0π = 2
0
sin3 θ dθ.
Ignoring the constants for the moment, consider
π
A similar reduction formula to (∗) can be obtained for
if
π/2
cosn x dx then Jn =
Jn =
0
cosn x dx (see exercises).. In particular
(n−1)
Jn−2
n
i.e. Wallis’ formula is the same for cosn x as for sinn x.
4. Harder Trigonometric Integrals
The following seemingly innocent integrals are examples, important in engineering, of trigonometric integrals that cannot be evaluated as indefinite integrals:
(a)
2
sin(x ) dx
and
cos(x2 ) dx
These are called Fresnel integrals.
sin x
dx
(b)
x
This is called the Sine integral.
Definite integrals of this type, which are what normally arise in applications, have to be evaluated by approximate numerical methods.
Fresnel integrals with limits arise in wave and antenna theory and the Sine integral with limits
in filter theory.
It is useful sometimes to be able to visualize the definite integral. For example consider
t
sin x
dx
t>0
F (t) =
x
0
11
HELM (VERSION 1: March 18, 2004): Workbook Level 1
13.6: Integration of Trigonometric Functions
Clearly, F (0) =
0
0
sin x
sin x
dx = 0. Recall the graph of
against x, x > 0:
x
x
sin x
x
t π
2π
x
For any positive value of t, F (t) is the shaded area shown (the area interpretation of a definite
integral was covered earlier in this Workbook). As t increases from 0 to π, it follows that F (t)
increases from 0 to a maximum value
π
sin x
F (π) =
dx
x
0
whose value could be determined numerically (it is actually about 1.85). As t further increases
sin x
curve
from π to 2π the value of F (t) will decrease to a local minimum at 2π because the
x
is below the x-axis between π and 2π.
Continuing to argue in this way we can obtain the shape of the F (t) graph as follows: (can you
see why the oscillations decrease in amplitude?)
F (t)
1.85
π
2
π
2π
t
The result
∞
sin x
π
dx =
x
2
0
is clear from the graph (you are not expected to know how this result is obtained). Such problems
are dealt with in Workbook 31.
HELM (VERSION 1: March 18, 2004): Workbook Level 1
13.6: Integration of Trigonometric Functions
12
Exercises
You will need to refer to a Table of Trigonometric Identities to answer these questions.
π/2
1. Find a) cos2 xdx b) 0 cos2 tdt c) (cos2 θ + sin2 θ)dθ
2. Use the
identity
sin(A
+
B)
+
sin(A
−
B)
=
2
sin
A
cos
B
to
find
sin 3x cos 2xdx
3. Find (1 + tan2 x)dx.
4. The mean square value of a function f (t) over the interval t = a to t = b is defined to be
1
b−a
b
(f (t))2 dt
a
Find the mean square value of f (t) = sin t over the
interval t = 0 to t = 2π.
5(a) Show that the reduction formula for Jn = cosn x dx is
1
(n − 1)
cosn−1 (x) sin x +
Jn−2
n
n
(b) Using the above reduction formula show that
1
4
8
cos2 x sin x +
sin x
cos5 x dx = cos4 x sin x +
5
15
15
Jn =
(c) Show that if
π/2
n−1
n
Jn =
cos x dx then Jn =
Jn−2 (Wallis’ formula).
n
0
(d) Using Wallis’ formula show that
π/2
5
cos6 x dx = π.
32
0
Answers
4. 21 .
1
1. a) 21 x + 41 sin 2x + c. b). π/4. c). θ + c. 2. − 10
cos 5x − 21 cos x + c. 3. tan x + c.
13
HELM (VERSION 1: March 18, 2004): Workbook Level 1
13.6: Integration of Trigonometric Functions