A lecture on trigonometric integrals and trigonometric substitution

A lecture on trigonometric integrals and trigonometric substitution

1

Trigonometric Integrals

First, we must recall a few

trigonometric identities

:

sin

2

⁡

x

+

cos

2

⁡

x

=

1

(1)

sec

2

⁡

x

=

1

+

tan

2

⁡

x

(2)

sin

2

⁡

x

=

1

-

cos

⁡

(

2

⁢

x

)

2

(3)

cos

2

⁡

x

=

1

+

cos

⁡

(

2

⁢

x

)

2

(4)

sin

⁡

(

2

⁢

x

)

=

2

⁢

sin

⁡

x

⁢

cos

⁡

x

(5)

cos

⁡

(

2

⁢

x

)

=

cos

2

⁡

x

-

sin

2

⁡

x

.

(6)

The most usual

integrals

which involve

trigonometric functions

can
be solved using the identities above.

Example 1.1

.

∫

sin

⁡

x

⁢

d

⁢

x

=

-

cos

⁡

x

+

C

and

∫

cos

⁡

x

⁢

d

⁢

x

=

sin

⁡

x

+

C

are
immediate integrals.

Example 1.2

.

For

∫

sin

2

⁡

x

⁢

d

⁢

x

,

∫

cos

2

⁡

x

⁢

d

⁢

x

we use formulas (3) and
(4) respectively, e.g.

∫

sin

2

⁡

x

⁢

d

⁢

x

=

∫

1

-

cos

⁡

(

2

⁢

x

)

2

⁢

𝑑

x

=

1

2

⁢

∫

(

1

-

cos

⁡

(

2

⁢

x

)

)

⁢

𝑑

x

=

1

2

⁢

(

x

-

sin

⁡

(

2

⁢

x

)

2

)

+

C

.

Example 1.3

.

For integrals of the form

∫

cos

m

⁡

x

⁢

sin

⁡

x

⁢

d

⁢

x

or

∫

sin

m

⁡

x

⁢

cos

⁡

x

⁢

d

⁢

x

we use substitution with

u

=

cos

⁡

x

or

u

=

sin

⁡

x

respectively, e.g.

∫

cos

2

x

sin

x

d

x

=

∫

-

u

2

d

u

=

-

u

3

3

+

C

=

-

cos

3

⁡

x

3

+

C

.

[

u

=

cos

x

,

d

u

=

-

sin

x

d

x

]

In the following examples, we use equations (1) in the forms

sin

2

⁡

x

=

1

-

cos

2

⁡

x

or

cos

2

⁡

x

=

1

-

sin

2

⁡

x

to transform the
integral into one of the type described in Example

1.3

.

Example 1.4

.

∫

sin

3

⁡

x

⁢

d

⁢

x

=

∫

sin

2

⁡

x

⁢

sin

⁡

x

⁢

d

⁢

x

=

∫

(

1

-

cos

2

⁡

x

)

⁢

sin

⁡

x

⁢

d

⁢

x

=

∫

sin

⁡

x

⁢

d

⁢

x

-

∫

cos

2

⁡

x

⁢

sin

⁡

x

⁢

d

⁢

x

=

-

cos

⁡

x

+

cos

3

⁡

x

3

+

C

.

Similarly one can solve

∫

cos

3

⁡

x

⁢

d

⁢

x

.

Example 1.5

.

∫

cos

3

⁡

x

⁢

sin

2

⁡

x

⁢

d

⁢

x

=

∫

cos

2

⁡

x

⁢

cos

⁡

x

⁢

sin

2

⁡

x

⁢

d

⁢

x

=

∫

(

1

-

sin

2

⁡

x

)

⁢

cos

⁡

x

⁢

sin

2

⁡

x

⁢

d

⁢

x

=

∫

cos

⁡

x

⁢

sin

2

⁡

x

⁢

d

⁢

x

-

∫

cos

⁡

x

⁢

sin

4

⁡

x

⁢

d

⁢

x

=

sin

3

⁡

x

3

-

sin

5

⁡

x

5

+

C

.

Example 1.6

.

In order to solve

∫

cos

5

⁡

x

⁢

sin

3

⁡

x

⁢

d

⁢

x

we express it first
as

∫

cos

5

⁡

x

⁢

sin

2

⁡

x

⁢

sin

⁡

x

=

∫

cos

5

⁡

x

⁢

(

1

-

cos

2

⁡

x

)

⁢

sin

⁡

x

⁢

d

⁢

x

and then proceed as in the previous example.

One can use similar tricks to solve integrals which involve
products of powers of

sec

⁡

x

and

tan

⁡

x

, by using Equation
(2). Also, recall that the

derivative

of

tan

⁡

x

is

sec

2

⁡

x

while the derivative of

sec

⁡

x

is

sec

⁡

x

⁢

tan

⁡

x

.

Example 1.7

.

∫

tan

5

⁡

x

⁢

sec

4

⁡

x

⁢

d

⁢

x

=

∫

tan

5

⁡

x

⁢

sec

2

⁡

x

⁢

sec

2

⁡

x

⁢

d

⁢

x

=

∫

tan

5

⁡

x

⁢

(

1

+

tan

2

⁡

x

)

⁢

sec

2

⁡

x

⁢

d

⁢

x

=

∫

tan

5

⁡

x

⁢

sec

2

⁡

x

⁢

d

⁢

x

+

∫

tan

7

⁡

x

⁢

sec

2

⁡

x

⁢

d

⁢

x

=

tan

6

⁡

x

6

+

tan

8

⁡

x

8

+

C

.

Example 1.8

.

∫

tan

3

⁡

x

⁢

sec

4

⁡

x

⁢

d

⁢

x

=

∫

tan

⁡

x

⁢

tan

2

⁡

x

⁢

sec

4

⁡

x

⁢

d

⁢

x

=

∫

tan

⁡

x

⁢

(

sec

2

⁡

x

-

1

)

⁢

sec

4

⁡

x

⁢

d

⁢

x

=

∫

tan

⁡

x

⁢

sec

⁡

x

⁢

sec

5

⁡

x

⁢

d

⁢

x

-

∫

tan

⁡

x

⁢

sec

⁡

x

⁢

sec

3

⁡

x

⁢

d

⁢

x

=

sec

6

⁡

x

6

-

sec

4

⁡

x

4

+

C

.

2

Trigonometric Substitutions

One can easily deduce that

∫

0

1

1

-

x

2

⁢

𝑑

x

has value

π

4

. Why? Simply because the graph of the

function

y

=

1

-

x

2

is half a circumference of radius

r

=

1

(because
if you square both sides of

y

=

1

-

x

2

you obtain

x

2

+

y

2

=

1

which is the equation of a circle or radius

r

=

1

).
Therefore, the area under the graph is a quarter of the area of a
circle.

How does one compute

∫

0

1

1

-

x

2

⁢

𝑑

x

without using the

geometry

of the problem? This is the prototype of integral where a
trigonometric substitution will work very nicely. Notice that
neither substitution nor

integration by parts

will work
appropriately.

Example 2.1

.

Suppose we want to solve

∫

0

1

1

-

x

2

⁢

𝑑

x

with

analytic

methods. We will use a substitution

x

=

sin

⁡

θ

(so

θ

will be our new

variable

of integration), because, as we know from
Equation (1),

1

-

x

2

=

1

-

sin

2

⁡

θ

=

cos

⁡

θ

,
thus getting rid of the pesky

square root

. Notice that

d

⁢

x

=

cos

⁡

θ

⁢

d

⁢

θ

. We also need to find the new limits of
integration with respect to the new variable of integration,
namely

θ

. When

x

=

0

=

sin

⁡

θ

we must have

θ

=

0

.
Similarly, when

x

=

1

=

sin

⁡

θ

one has

θ

=

π

/

2

. We are
now ready to

integrate

:

∫

0

1

1

-

x

2

⁢

𝑑

x

=

∫

0

π

/

2

(

cos

⁡

θ

)

⁢

cos

⁡

θ

⁢

d

⁢

θ

=

∫

0

π

/

2

cos

2

⁡

θ

⁢

d

⁢

θ

=

∫

0

π

/

2

1

+

cos

⁡

(

2

⁢

θ

)

2

⁢

𝑑

θ

=

1

2

⁢

(

θ

+

sin

⁡

(

2

⁢

θ

)

2

)

0

π

/

2

=

π

/

4

.

Notice that we made use of Equation (4) in the second line.

Example 2.2

.

Similarly, one can solve

∫

0

r

r

2

-

x

2

⁢

𝑑

x

by using a
substitution

x

=

r

⁢

sin

⁡

θ

. Indeed,

r

2

-

x

2

=

r

2

-

r

2

⁢

sin

2

⁡

θ

=

r

⁢

cos

⁡

θ

and

d

⁢

x

=

r

⁢

cos

⁡

θ

⁢

d

⁢

θ

. The limits of integration with respect
to

θ

are again

θ

=

0

to

θ

=

π

/

2

(check this!).
Thus:

∫

0

r

r

2

-

x

2

⁢

𝑑

x

=

∫

0

π

/

2

r

2

⁢

(

cos

⁡

θ

)

⁢

cos

⁡

θ

⁢

d

⁢

θ

=

r

2

⁢

∫

0

π

/

2

cos

2

⁡

θ

⁢

d

⁢

θ

=

r

2

⁢

∫

0

π

/

2

1

+

cos

⁡

(

2

⁢

θ

)

2

⁢

𝑑

θ

=

r

2

2

⁢

(

θ

+

sin

⁡

(

2

⁢

θ

)

2

)

0

π

/

2

=

r

2

⁢

π

/

4

.

Thus, we have proved that a quarter of a circle of radius

r

has
area

r

2

⁢

π

/

4

which implies that the area of such a circle is

π

⁢

r

2

, as usual.

The trigonometric substitutions

usually

work when

expressions

like

r

2

-

x

2

,

r

2

+

x

2

,

x

2

-

r

2

appear in the integral at hand, for some

real
number

r

. Here is a table of the suggested change of variables
in each particular case:

If you see…

try this…

because…

1

-

x

2

x

=

sin

⁡

θ

1

-

sin

2

⁡

θ

=

cos

⁡

θ

r

2

-

x

2

x

=

r

⁢

sin

⁡

θ

r

2

-

sin

2

⁡

θ

=

r

⁢

cos

⁡

θ

1

+

x

2

x

=

tan

⁡

θ

1

+

tan

2

⁡

θ

=

sec

⁡

θ

r

2

+

x

2

x

=

r

⁢

tan

⁡

θ

r

2

+

tan

2

⁡

θ

=

r

⁢

sec

⁡

θ

x

2

-

1

x

=

sec

⁡

θ

sec

2

⁡

θ

-

1

=

tan

⁡

θ

x

2

-

r

2

x

=

r

⁢

sin

⁡

θ

sec

2

⁡

θ

-

1

=

r

⁢

tan

⁡

θ

Remark 2.3

.

The above are “suggested” substitutions, they may not be the
most ideal choice! For example, for the integral

∫

2

⁢

x

⁢

1

-

x

2

⁢

𝑑

x

, the change

u

=

1

-

x

2

will work much better than

x

=

sin

⁡

θ

.

Example 2.4

.

We would like to find the value of

∫

2

2

1

x

3

⁢

x

2

-

1

⁢

𝑑

x

.

Since neither a

u

-substitution nor integration by parts seem
appropriate, we try

x

=

sec

⁡

θ

,

d

⁢

x

=

sec

⁡

θ

⁢

tan

⁡

θ

⁢

d

⁢

θ

. When

x

=

2

=

sec

⁡

θ

one has

θ

=

π

/

4

while

x

=

2

implies

θ

=

π

/

3

. Hence:

∫

2

2

1

x

3

⁢

x

2

-

1

⁢

𝑑

x

=

∫

π

/

4

π

/

3

sec

⁡

θ

⁢

tan

⁡

θ

sec

3

⁡

θ

⁢

tan

⁡

θ

⁢

𝑑

θ

=

∫

π

/

4

π

/

3

1

sec

2

⁡

θ

⁢

𝑑

θ

=

∫

π

/

4

π

/

3

cos

2

⁡

θ

⁢

d

⁢

θ

and the last integral is easy to compute using Equation (4).

Title

A

lecture on trigonometric integrals and trigonometric substitution

Canonical

name

ALectureOnTrigonometricIntegralsAndTrigonometricSubstitution

Date of creation

2013-03-22 15:38:39

Last modified on

2013-03-22 15:38:39

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Feature

Classification

msc 26A36

Related topic

ALectureOnIntegrationByParts

Related topic

ALectureOnIntegrationBySubstitution

Related topic

ALectureOnThePartialFractionDecompositionMethod

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