Trigonometric Substitutions & Trigonometric Integrals
Big Picture. This topic has two halves that connect:
- Trigonometric integrals — integrating expressions that already contain $\sin$, $\cos$, $\tan$, $\sec$, etc., by rewriting them with identities.
- Trigonometric substitution — introducing a trig function into an algebraic integrand to remove a square-root expression.
1. Powers of Sine and Cosine: $\displaystyle\int \sin^m x\,\cos^n x\,dx$
The strategy depends entirely on the parity (odd/even) of the exponents $m$ and $n$.
Case A — At Least One Odd Power
Rule: Strip off one copy of the function with the odd power to pair with $dx$, convert the remaining even power using the Pythagorean identity, and substitute.
| Situation | Strip | Identity | Substitution |
|---|---|---|---|
| $n$ (cosine power) is odd | one $\cos x\,dx$ | $\cos^2 x = 1-\sin^2 x$ | $u=\sin x,\;du=\cos x\,dx$ |
| $m$ (sine power) is odd | one $\sin x\,dx$ | $\sin^2 x = 1-\cos^2 x$ | $u=\cos x,\;du=-\sin x\,dx$ |
| Both odd | Either one — pick whichever is simpler | Same as above | Same as above |
Quick example (odd cosine power):
$$\int \sin^2 x\,\cos^3 x\,dx = \int \sin^2 x\,(1-\sin^2 x)\,\cos x\,dx$$Set $u=\sin x$:
$$= \int u^2(1-u^2)\,du = \int (u^2-u^4)\,du = \frac{u^3}{3}-\frac{u^5}{5}+C = \frac{\sin^3 x}{3}-\frac{\sin^5 x}{5}+C$$Case B — Both Powers Even
Rule: Use the power-reduction (half-angle / double-angle) formulas repeatedly:
$$\sin^2 x = \frac{1-\cos 2x}{2}, \qquad \cos^2 x = \frac{1+\cos 2x}{2}$$