Mathematics 136 – AP Calculus Summary of Trigonometric Substitutions Fall 2009

The trigonometric substitution method handles many integrals containing expressions like

\sqrt{a^{2} − x^{2}}, \sqrt{x^{2} + a^{2}}, \sqrt{x^{2} − a^{2}}

(possibly including expressions without the square roots!) The basis for this approach is the trigonometric identities

1 = sin^{2} θ + cos^{2} θ
⇒ sec^{2} θ = tan^{2} θ + 1.

from which we derive other related identities:

\sqrt{a^{2} − (a sin θ)^{2}} = a cos θ
\sqrt{(a tan θ)^{2} + a^{2}} = a sec θ
\sqrt{(a sec θ)^{2} − a^{2}} = a tan θ

Hence,

1. If our integral contains \sqrt{a^{2} − x^{2}}, the substitution x = a sin θ will convert this radical to the simpler form a cos θ.

2. If our integral contains \sqrt{x^{2} + a^{2}}, the substitution x = a tan θ will convert this radical to the simpler form a sec θ.

3. If our integral contains \sqrt{x^{2} − a^{2}}, the substitution x = a sec θ will convert this radical to the simpler form a tan θ.

Then we substitute for the rest of the integral, integrate the resulting trigonometric form, and convert back to the original variable.

Two Examples

A) Compute ∫ \frac{u^{2}}{\sqrt{a^{2}−u^{2}}} du. Solution: The \sqrt{a^{2} − u^{2}} tells us that we want the sine substitution: u = a sin θ. Then du = a cos θ dθ, and the integral becomes:

∫ \frac{a^{3} sin^{2} θ cos θ dθ}{a cos θ} = a^{2} ∫ sin^{2} θ dθ

We apply the reduction formula for powers of sin (this is # 73 in Stewart’s table of integrals from the Reference Pages handout from last week):

a^{2} ∫ sin^{2} θ dθ = \frac{−a^{2}}{2} sin θ cos θ + \frac{a^{2}}{2} ∫ dθ = \frac{−a^{2}}{2} sin θ cos θ + \frac{a^{2}}{2} θ + C.

Then, we convert back to functions of u using the substitution equation u = a sin θ. From this,

θ = arcsin(u/a),  cos θ = \sqrt{a^{2} − u^{2}}/a,  sin θ = u/a.

So the integral is

∫ \frac{u^{2}}{\sqrt{a^{2} − u^{2}}} du = \frac{−1}{2}u\sqrt{a^{2} − u^{2}} + \frac{a^{2}}{2} arcsin(u/a) + C.

Note that the final answer here matches # 34 in the table of integrals.

B) Compute ∫ \frac{dx}{x\sqrt{x^{2}+16}}. Solution: The \sqrt{x^{2} + 16} indicates that we want the tangent substitution x = 4 tan θ. Then dx = 4 sec^{2} θ dθ and the integral becomes:

∫ \frac{4 sec^{2} θ dθ}{4 tan θ · 4 sec θ} = \frac{1}{4} ∫ \frac{sec θ}{tan θ} dθ = \frac{1}{4} ∫ \frac{1}{sin θ} dθ

Since 1/ sin(θ) = csc(θ), from # 15 in Stewart’s table of integrals,

\frac{1}{4} ∫ csc(θ) dθ = \frac{1}{4} ln |csc(θ) − cot(θ)| + C

Then, from x = 4 tan θ, we get cos θ = \frac{4}{\sqrt{x^{2}+16}} and sin θ = \frac{x}{\sqrt{x^{2}+16}}. Hence

csc(θ) = \frac{\sqrt{x^{2} + 16}}{x} and cot(θ) = \frac{4}{x}.

The integral equals:

\frac{1}{4} ln |\frac{\sqrt{x^{2} + 16}}{x} − \frac{4}{x}| + C.