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Integration using
trig identities or
a trig substitution
Some integrals involving trigonometric functions can be evaluated by using the trigonometric
identities
...

On occasions a trigonometric substitution will enable an integral to be evaluated
...

In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature
...

• integrate products of sines and cosines using a mixture of trigonometric identities and
integration by substitution
• use trigonometric substitutions to evaluate integrals

Contents
1
...
Integrals requiring the use of trigonometric identities

2

3
...
Integrals which make use of a trigonometric substitution

1

2

5

c mathcentre August 28, 2004

1
...
In case like these trigonometric

identities can be used to write the integrand in an alternative form which can be integrated
more readily
...
Such substitutions are described in Section 4
...
Integrals requiring the use of trigonometric identities
The trigonometric identities we shall use in this section, or which are required to complete the
Exercises, are summarised here:
2 sin A cos B
2 cos A cos B
2 sin A sin B
2
sin A + cos2 A
cos 2A

=
=
=
=
=
=
=
sin 2A =
1 + tan2 A =

sin(A + B) + sin(A − B)
cos(A − B) + cos(A + B)
cos(A − B) − cos(A + B)
1
cos2 A − sin2 A
2 cos2 A − 1
1 − 2 sin2 A
2 sin A cos A
sec2 A

Some commonly needed trigonometric identities

Example

π

sin2 x dx
...
The trigonometric identity we shall use here is one of
the ‘double angle’ formulae:
cos 2A = 1 − 2 sin2 A
By rearranging this we can write

1
sin2 A = (1 − cos 2A)
2
Notice that by using this identity we can convert an expression involving sin2 A into one which
has no powers in
...


Note that the integrand is a product of the functions sin 3x and cos 2x
...

With A = 3x and B = 2x we have
1
2
1
=
2

sin 3x cos 2x dx =

(sin 5x + sin x)dx

1
− cos 5x − cos x + c
5
1
1
= − cos 5x − cos x + c
10
2
Exercises 1
Use the trigonometric identities stated on page 2 to find the following integrals
...
(a)

2

cos x dx

cos2 x dx

(b)

(c)

sin 2x cos 2x dx

0
π/3

2
...


π/6

3
...
In the first example we see

how to deal with integrals in which m is odd
...


Study of the integrand, and the table of identities shows that there is no obvious identity which
will help us here
...
The reason for doing this will become apparent
...

Example
To find

sin4 x cos3 x dx we write

sin4 x(cos2 x·cos x) dx
...

Exercises 2
1
...
Evaluate

sin2 x cos2 xdx by using the double angle formulae

(b)

sin2 x =

cos5 x dx

1 − cos 2x
2

3
...


cos2 x =

1 + cos 2x
2

sin4 x cos2 x dx
...
Integrals which make use of a trigonometric substitution
There are several integrals which can be found by making a trigonometric substitution
...

Example
1
dx
...

Suppose we wish to find

Our reason for doing this is that the integrand will then involve

1
and we have an
1 + tan2 θ

identity (1 + tan2 A = sec2 A) which will enable us to simplify this
...
The integral becomes
dθ
1
1
dx =
sec2 θ dθ
2
1+x
1 + tan2 θ
1
sec2 θ dθ
=
2θ
sec
=

1 dθ

= θ+c
= tan−1 x + c
1
dx = tan−1 x + c
...

2
1+x
1
dx:
We can generalise this result to the integral
2 + x2
a
We make the substitution x = a tan θ, dx = a sec2 θ dθ
...

1
a

1 dθ =

Key Point
1
dx = tan−1 x + c
1 + x2

5

a2

1
1
x
dx = tan−1 + c
2
+x
a
a

c mathcentre August 28, 2004

Example
Suppose we seek

1
dx
...
Let us observe the effect of making the substitution u = 2 x, so that u = 4 x2
...

6
6
2

We now consider a similar example for which a sine substitution is appropriate
...

a2 − x2
The substitution we will use here is based upon the observations that in the denominator we
have a term a2 − x2 , and that there is a trigonometric identity 1 − sin2 A = cos2 A (and hence
(a2 − a2 sin2 A = a2 cos2 A)
...
Then
becomes
√

1
dx =
a2 − x2
=
=
=

dx
dθ

= a cos θ and dx = a cos θdθ
...

2
a
−x
This is another standard result
...

4 − 9x2
The trick is to try to write this in the standard form
...
Use the trigonometric substitution indicated to find the given integral
...

dx let x = 4 sin θ
(b)
2
2
2
1 + 4x
16 − x
Answers
Exercises 1
cos 4x
x 1
(c) −
1
...
165 (3 d
...
)
(b) t + c
2
...
(a) 3 cos2 x sin x + 2 sin x + c
(b) 1 cos4 x sin x + 15 cos2 x sin x + 15 sin x + c
3
5
1
4
8
(c) − 7 sin4 x cos3 x − 35 sin2 x cos3 x − 105 cos3 x + c
...
− 4 sin x cos3 x + 1 cos x sin x + 8 x + c
...
− 6 sin3 x cos3 x − 1 sin x cos3 x + 16 cos x sin x +
8

Exercises √
3
1
1
...


c mathcentre August 28, 2004



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