Trigonometric Substitution

How to solve the given integral by using the trigonometric substitution: 2 examples and their solutions.

Example 1

Example

Solution

If an integral has a2 - x2, especially √a2 - x2,
then set x = a sin θ.

Differentiate both sides.

Then dx = a cos θ dθ.

Derivative of a Composite Function

Derivative of sin x

Before using x and dx,
write an expression for θ.

Start from x = a sin θ.
sin θ = x/a
So θ = arcsin x/a.

Arcsine: Value

x = a sin θ
dx = a cos θ dθ

Put these into the given integral.

Then (given) = ∫ [a cos θ]/√a2 - a2 sin2 θ dθ.

Integral by Substitution: Indefinite Integral

a2 - a2 sin2 θ = a2(1 - sin2 θ)

Common Monomial Factor

1 - sin2 θ = cos2 θ

Pythagorean Identity

a2 cos2 θ = a cos θ

Simplify a Radical

∫ [a cos θ]/[a cos θ] dθ
= ∫ 1 dθ
= ∫ dθ

∫ dθ = θ + C

Integral of a Polynomial

Change θ back to x.

θ = arcsin x/a

So θ + C = arcsin x/a + C.

So arcsin x/a + C is the answer.

Example 2

Example

Solution

If an integral has a2 + x2,
then set x = a tan θ.

Differentiate both sides.

Then dx = a sec2 θ dθ.

Derivative of tan x

Before using x and dx,
write an expression for θ.

Start from x = a tan θ.
tan θ = x/a
So θ = arctan x/a.

Arctangent: Value

a tan θ
a sec2 θ

Put these into the given integral.

Then (given) = ∫ [a sec2 θ]/[a2 + a2 tan2 θ] dθ.

a2 + a2 tan2 θ = a2(1 + tan2 θ)

1 + tan2 θ = sec2 θ

Pythagorean Identity

∫ [a sec2 θ]/[a2 sec2 θ] dθ = ∫ 1/a dθ

Take 1/a out from the integral.

[1/a]∫ dθ = [1/a]θ + C

Change θ back to x.

θ = arctan x/a

So [1/a]θ + C = [1/a]arctan x/a + C.

So [1/a]arctan x/a + C is the answer.