Trigonometric Substitution

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

Example 1: a2 - x2

For a2 - x2 form,
set x = a sin θ.

Then dx = a cos θ .

Derivative of sin x

And from x = a sin θ,
find θ: arcsin (x/a).

Soving arcsine functions

Substitute [x] with [a sin θ].
And substitute [dx] with [a cos θ ].

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

Integration by substitution (Part 1)

Take a cos θ out from the square root.

Cancel a cos θ
on both of the numerator and the denominator.

Integrate .

Then (given) = θ + C.

Antiderivative

Put arcsin (x/a) in θ.

Then (given) = arcsin (x/a) + C.

Example 2: a2 + x2

For a2 + x2 form,
set x = a tan θ.

Then dx = a sec2 θ .

Derivative of tan x

And from x = a tan θ,
find θ: arctan (x/a).

Soving arctangent functions

Substitute [x] with [a tan θ].
And substitute [dx] with [a sec2 θ ].

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

Integration by substitution (Part 1)

Cancel a sec2 θ
on both of the numerator and the denominator.

1/a is a coefficient.

So (given) = (1/a)θ + C.

Put arctan (x/a) in θ.

Then (given) = (1/a) arctan (x/a) + C.

Example 3: x2 - a2

For x2 - a2 form,
set x = a sec θ.

Then dx = a sec θ tan θ .

Derivative of sec x

And from x = a sec θ,
sec θ = x/a.

Substitute [x] with [a sec θ].
And substitute [dx] with [a sec θ tan θ ].

Then (given) = ∫ (a sec θ tan θ)/(a2 sec2 θ - a2) .

Integration by substitution (Part 1)

1 + tan2 θ = sec2 θ

So sec2 θ - 1 = tan2 θ.

Pythagorean identities

Take (a tan θ) out from the square root.

Cancel (a tan θ)
on both of the numerator and the denominator.

Integrate sec θ .

Then (given) = ln |sec θ + tan θ| + C.

Indefinite integration of sec x

You know that sec θ = x/a.
So find tan θ.

Draw a right triangle
that satisfies sec θ = x/a.

Cosine: CAH
So sec θ = (hypotenuse)/(adjacent side)
= x/a

Trigonometric ratio - Secant

(hypotenuse) = x