Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
INTEGRATION WITH TRIG
IDENTITIES
CALCULUS 9
INU0115/515 (M ATHS 2)
Dr Adrian Jannetta MIMA CMath FRAS
Integration with trig identities
1 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Objectives
In this presentation we’re going to tackle integration of trig functions.
Before you begin:
• You should be comfortable doing integration by reverse chain rule.
• A list of common trig identities will also prove to be useful!
The strategy for the problems you’re about to see is to replace the
function with something we can integrate easily (such as a function on
our list of standard integrals). There are many trig identities in
mathematics but we’ll begin with some simple examples to cover the
most common cases you’re likely to see.
Integration with trig identities
2 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Integrating tan2 x
Let’s start with a simple example...
Using identities
Z
Find
tan2 x dx
We can use the identity tan2 x ≡ sec2 −1 in this case:
Z
Z
tan2 x dx =
(sec2 x − 1) dx
The integral of sec2 x is on our list of standard integrals:
Z
Z
tan2 x dx =
Integration with trig identities
(sec2 x − 1) dx = tan x − x + C
3 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Integrating sin2 x and cos2 x
Double angle identities can be useful for integrating sin2 x or cos2 x.
Using a double angle identity
Find
Z
sin2 x dx.
We can use the identity sin2 θ ≡ 21 (1 − cos 2θ ) to give us an expression that can be
integrated more easily.
Z
Z
sin2 x dx =
1
2
(1 − cos 2x) dx
Using the reverse chain rule on the cosine term, we can write:
Z

‹
sin 2x
sin2 x dx = 12 x −
+C
2
=
1
4 (2x − sin 2x) + C
Higher powers of sine or cosine (sink x or cosk x, k > 2) will be examined later in
the presentation but will not be assessed on the exam.
Integration with trig identities
4 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Double angle indentities
Integration with a double angle identity
Evaluate
Z
π
12
8 sin 2x cos 2x dx
0
We’ll need the identity sin 2θ ≡ 2 sin θ cos θ . In this case, we have θ = 2x so that:
sin 4x
≡
2 sin 2x cos 2x
∴ 4 sin 4x
≡
8 sin 2x cos 2x
This integral becomes:
Z
π
12
8 sin 2x cos 2x dx =
0
Integrate with the reverse chain rule:
Z π
12
∴
=
[− cos 4x]0
8 sin 2x cos 2x dx
=
1
2
π
12
π
12
4 sin 4x dx
0
π/12
8 sin 2x cos 2x dx
0
Z
Z
= − cos π3 − (− cos 0) = − 21 + 1
0
Integration with trig identities
5 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Using compound angle identities
Recall the following identities for sine:
sin(A + B)
sin(A − B)
≡
sin A cos B + cos A sin B
≡
sin A cos B − cos A sin B
≡
cos A cos B − sin A sinB
and for cosine:
cos(A + B)
cos(A − B)
≡
cos A cos B + sin A sinB
We’re going to need these for the next couple of examples.
Compound angle identities are useful for turning a product into a sum; if
the integral is a sum of two terms then we can integrate each term
without a problem.
Integration with trig identities
6 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Using compound angle identities
(a) Show that sin A cos B ≡ 21 [sin(A + B) + sin(A − B)]
(b) Hence, evaluate the integral
Z
π
12
sin 6x cos 3x dx
0
(a) Use the identies for sin(A + B) and sin(A − B)
sin(A + B)
sin(A − B)
≡
≡
sin A cos B + sin B cos A
sin A cos B − sin B cos A
Add the identities:
sin(A + B) + sin(A − B) ≡ 2 sin A cos B
Divide both sides by 2 to get:
sin A cos B ≡ 21 [sin(A + B) + sin(A − B)]
Integration with trig identities
7 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
(b) To integrate the product sin6x cos 3x we’ll use the identity obtained
in part (a).
Set A = 6x and B = 3x.
Z
π
12
sin 6x cos 3x dx
0
π
12
≡
Z
=
1
2
=
1
− 18
[cos 9x + 3 cos 3x]012
3π
1
cos 9π
− 18
12 + 3 cos 12 − (cos 0 + 3 cos 0)
1
π
− 18
cos 3π
4 + 3 cos 4 − (1 + 3)
”€ p
—
p Š
1
− 18
− 22 + 3 2 2 − 4
p
1
( 2 − 4)
− 18
1
2
[sin9x + sin 3x] dx
0
1
π
− 9 cos 9x − 13 cos 3x 012
π
=
=
=
=
Integration with trig identities
8 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
How can we figure out which compound angle identity to use?
Here’s the list again:
sin(A + B)
=
sin A cos B + cos A sinB
(1)
sin(A − B)
=
sin A cos B − cos A sinB
(2)
cos(A + B)
=
(3)
cos A cos B + sin A sinB
(4)
R
If we have to integrate a product such as cos 6x cos 2x dx then we’d use
the last pair of identities. because they contain the parts we need on the
LHS.
cos(A − B)
Integration with trig identities
=
cos A cos B − sin A sinB
9 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Using compound angle identities
Find
Z
cos 6x cos 2x dx.
We can derive the identity we need from the cosine compound angle identities
(equations 3 and 4):
cos(A + B) + cos(A − B) = 2 cos A cosB
Therefore
cos A cos B = 12 [cos(A + B) + cos(A − B)]
That means we can express the integrand like this:
Z
Z
cos 6x cos 2x dx
=
1
2
=
1
2
=
Integration with trig identities
[cos 8x + cos 4x] dx
‹
sin 8x sin 4x
+C
+
8
4
1
16 (sin 8x + 2 sin 4x) + C

10 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Powers of sine, cosine and tangent (optional)
The following methods are optional and will not be assessed on the
exam.
Trig identities are essential for integrating powers
of sine, cosine or
R
tangent. For example, we may wish to find cos5 x dx. The general
procedure is decided by whether the power is odd or even (for sine and
cosine) or a power of the tangent function.
1
Even powers of sin x and cos x.
Use double angle identities.
1
2
Odd powers of sin x and cos x.
Use cos2 x ≡ 1 − sin2 x or sin2 x ≡ 1 − cos2 x.
Powers of tan x.
Use tan2 x ≡ sec2 x − 1.
We’ll illustrate each of these cases with an example.
Integration with trig identities
11 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Even powers of sine and cosine
Z
Find
cos4 x dx
Separate the integrand into powers of cos2 x and expand with the double
angle identity:
Z
Z
2
4
cos2 x dx
cos x dx =
=
Z
1
2 (1 + cos 2x)
=
1
4
Z
=
1
4
Z
2
dx
(1 + cos 2x)2 dx
(1 + 2 cos 2x + cos2 2x) dx
We must apply the double angle identity again to remove the remaining
powers.
Integration with trig identities
12 / 18
Adrian Jannetta
Introduction
Double angle identities
Z
Compound angle identities
4
cos x dx =
1
4
Z
Higher powers
Summary
Test
(1 + 2 cos 2x + cos2 2x) dx
This time put cos2 2x ≡ 12 (1 + cos 4x):
Z
=
1
4
=
1
4
Z
=
1
4
Z
1 + 2 cos 2x + 12 (1 + cos 4x) dx
1 + 2 cos 2x + 12 + 12 cos 4x dx
3
2
+ 2 cos 2x + 12 cos 4x dx
Now we can integrate the cosine terms:
2 sin 2x 12 sin4x
1 3
= 4 2x+
+
+C
2
4
=
∴
Integration with trig identities
3
1
8 x + sin 2x + 8 sin 4x + C
Z
cos4 x dx = 18 (3x + 8 sin 2x + sin 4x) + C
13 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Odd powers of sine and cosine
Find
Z
sin5 x dx
The first thing to do in this case is to ‘take out’ a factor of sin x
Z
Z
sin5 x dx =
sin x. sin4 x dx
and use the identity sin2 x ≡ 1 − cos2 x to introduce the cosine function. This is
necessary because it puts the integral into a form that we can integrate by
substitution.
Z
sin x(sin2 x)2 dx
=
Integration with trig identities
=
Z
sin x(1 − cos2 x)2 dx
=
Z
sin x(1 − 2 cos2 x + cos4 x) dx
=
Z
(sin x − 2 sin x cos2 x + sin x cos4 x) dx
14 / 18
Adrian Jannetta
Introduction
Double angle identities
Z
5
Compound angle identities
sin x dx =
Z
Higher powers
Summary
Test
(sin x − 2 sinx cos2 x + sin x cos4 x) dx
The first term can be integrated by immediately. The other terms are to
be integrated by substitution; put u = cos x in each case, so that
Z
2 cos3 x
2 sin x cos2 x dx = −
+ K1
3
where K1 is a constant, and
Z
sinx cos4 x dx = −
where K2 is another constant.
Z
∴
Integration with trig identities
sin5 x dx = − cos x +
cos5 x
+ K2
5
2 cos3 x cos5 x
−
+C
3
5
15 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Powers of tangent
Find
Z
tan4 x dx
Take a factor of tan2 x from the integrand:
Z
Z
tan4 x dx =
tan2 x tan2 x dx
Use the identity tan2 x ≡ sec2 x − 1:
Z
=
sec2 x − 1 tan2 x dx
Z
=
sec2 x tan2 x dx − tan2 x dx
Replace the second term using the identity
again:
Z
=
sec2 x tan2 x dx − (sec 2 x − 1) dx
=
Z
sec2 x tan2 x dx − sec 2 x + 1 dx
Integration with trig identities
Use the substitution u = tan x to integrate
the first term on the RHS.
Z
sec2 x tan2 x dx =
1
3
tan3 x + K1
The second term on the RHS can be
integrated by inspection:
Z
(− sec2 x + 1) dx = − tan x + x + K2
where K1 and K2 are constants.
Collecting these results we find:
Z
tan4 x dx =
16 / 18
1
3
tan3 x − tan x + x + C
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Summary
Integration can sometimes be simplified by substituting with a known
trig identity. The following identities are usually useful:
Double angle for sine:
sin 2x ≡ 2 sin x cos x
To integrate powers of sine or cosine:
sin2 x ≡ 12 (1 − cos 2x)
and cos2 x ≡ 21 (1 + cos 2x)
The integral of tan2 x can be found from
tan2 x ≡ sec2 x − 1
Compound angles are useful for turning products into sums:
sin(A ± B)
cos(A ± B)
≡
≡
sin A cos B ± cos A sinB
cos A cos B ∓ sin A sinB
This isn’t a complete list...you may find other identities helpful too!
Integration with trig identities
17 / 18
Adrian Jannetta
Introduction
Double angle identities
Compound angle identities
Higher powers
Summary
Test
Test yourself...
Use partial fractions to integrate the following. Write your answer in the simplest
possible form.
Z
cos2 2x dx
1
2
Z
3
Z
4
Z
π
12
tan2 3x dx
0
sin 4x cos 3x dx.
Hint: use sin(A + B) + sin(A − B).
π
6
16 sin x cos x dx
0
Answers:
1
2
3
1
1
2 x + 8 sin 4x + C
π
1
−
3
12
1
(cos 7x + 7 cos x) + C.
− 14
4 2.
Integration with trig identities
18 / 18
Adrian Jannetta