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Calculus II (Integration & Series) Unit 3: Trig integrals & trig substitution

Trig integrals use identities to reduce powers. Trig substitution uses right-triangle relationships to remove square roots.

Covers: Powers of sin/cos/tan, trig substitution

Study guide

Integrals of powers of trig functions use identities to set up a substitution. For ∫ sinᵐx cosⁿx dx, if one power is odd, save one factor of that function, convert the rest using sin²x + cos²x = 1, and substitute u for the other function. For example, ∫ sin³x dx: save sin x, write sin²x = 1 − cos²x, and let u = cos x. If both powers are even, use the half-angle identities sin²x = (1 − cos 2x)/2 and cos²x = (1 + cos 2x)/2 to reduce the powers.

Integrals with tangent and secant use 1 + tan²x = sec²x. With an even power of secant, save sec²x and let u = tan x; with an odd power of tangent and some secant, save sec x tan x and let u = sec x.

Trig substitution handles square roots of quadratics. For √(a² − x²), let x = a sin θ; for √(a² + x²), let x = a tan θ; for √(x² − a²), let x = a sec θ. Each substitution uses a Pythagorean identity to remove the root. Replace dx too (for x = a sin θ, dx = a cos θ dθ), integrate in θ, then draw a right triangle from the substitution to rewrite the answer in terms of x. For definite integrals you can change the limits to θ-values instead.

If the root contains a general quadratic such as √(x² + 2x + 5), complete the square first to get √((x + 1)² + 4). Let u = x + 1, then use the matching substitution, here u = 2 tan θ. For √(a² − b²x²), let x = (a/b) sin θ so the identity still removes the root. Similarly, use x = (a/b) tan θ for √(a² + b²x²) and x = (a/b) sec θ for √(b²x² − a²).

Tangent and secant substitutions often lead to ∫ sec θ dθ = ln|sec θ + tan θ| + C and ∫ sec³θ dθ = ½(sec θ tan θ + ln|sec θ + tan θ|) + C, both worth knowing. When the answer in θ contains sin 2θ (common after a half-angle step), rewrite it as 2 sin θ cos θ before using the reference triangle to return to x.

As an alternative for √(x² + a²) or √(x² − a²), let x = a sinh t or x = a cosh t, and use cosh²t − sinh²t = 1. Answers then come out in terms of inverse hyperbolic functions or logarithms.

Key ideas

• Odd power of sin or cos: save one factor and convert the rest with sin² + cos² = 1.
• Even powers of both: use half-angle identities.
• √(a² − x²) → x = a sin θ; √(a² + x²) → x = a tan θ; √(x² − a²) → x = a sec θ.
• After integrating, draw a triangle to convert back to x.

Formulas and rules

Common mistakes

• Forgetting dx = a cos θ dθ when substituting.
• Leaving the answer in θ.

Words to know

Trig substitution for √(a² − x²)

x = a sin θ

Trig substitution for √(a² + x²)

x = a tan θ

Half-angle identity for sin²x

(1 − cos 2x)/2

Odd power of sine

Save one sin x, rewrite sin²x = 1 − cos²x, let u = cos x.

Back-substitution (trig sub)

Use a right triangle to rewrite the answer in terms of x.

Trig substitution for √(x² − a²)

x = a sec θ

Even powers of sine and cosine

Use half-angle identities to reduce the powers.

Half-angle identity for cos²x

(1 + cos 2x)/2

Even power of secant

Save sec²x, convert the rest with 1 + tan²x = sec²x, let u = tan x.

Odd power of tangent

Save sec x tan x, convert to secants, let u = sec x.

dx in trig substitution

For x = a sin θ, dx = a cos θ dθ.

Sample practice questions

23 more questions like these, from level 1 to 5, are in the practice set.