Integrals like $\int \sin^{3}x\cos^{2}x\,dx$ or $\int \sin^{2}x\,dx$ come up in physics, in Fourier series and inside trigonometric substitution. Two tools handle almost all of them: substitution with $u = \sin x$ or $u = \cos x$, and the power-reducing identities.
Why it works
Substitution needs a spare factor to be $du$. If $\sin x$ has an odd power, peel off one $\sin x$ to go with $dx$ (since $d(\cos x) = -\sin x\,dx$), and the remaining even power becomes a polynomial in $\cos x$ through $\sin^{2}x = 1 - \cos^{2}x$. Then $u = \cos x$ leaves a polynomial in $u$.
If both powers are even, there is no factor to spare. Instead lower the powers with
$$\sin^{2}x = \frac{1 - \cos 2x}{2},$$ $$\cos^{2}x = \frac{1 + \cos 2x}{2},$$
which trade a square for a single cosine of a double angle.
Key idea
For $\int \sin^{m}x\cos^{n}x\,dx$: if $m$ is odd, save one $\sin x$ and let $u = \cos x$; if $n$ is odd, save one $\cos x$ and let $u = \sin x$; if both are even, use the power-reducing identities.
| Integrand | Save | Convert the rest with | Substitute |
|---|---|---|---|
| $\sin^{m}x\cos^{n}x$, $m$ odd | $\sin x\,dx$ | $\sin^{2}x = 1 - \cos^{2}x$ | $u = \cos x$ |
| $\sin^{m}x\cos^{n}x$, $n$ odd | $\cos x\,dx$ | $\cos^{2}x = 1 - \sin^{2}x$ | $u = \sin x$ |
| $\sin^{m}x\cos^{n}x$, both even | — | $\sin^{2}x = \frac{1 - \cos 2x}{2}$, $\cos^{2}x = \frac{1 + \cos 2x}{2}$ | — |
| $\tan^{m}x\sec^{n}x$, $n$ even | $\sec^{2}x\,dx$ | $\sec^{2}x = 1 + \tan^{2}x$ | $u = \tan x$ |
| $\tan^{m}x\sec^{n}x$, $m$ odd | $\sec x\tan x\,dx$ | $\tan^{2}x = \sec^{2}x - 1$ | $u = \sec x$ |
Worked example: An odd power of sine
Find $\int \sin^{3}x\cos^{2}x\,dx$.
- Save one $\sin x$: $\sin^{3}x = (1 - \cos^{2}x)\sin x$.
- Let $u = \cos x$, $du = -\sin x\,dx$: the integral becomes $-\int (1 - u^{2})u^{2}\,du = -\int (u^{2} - u^{4})\,du$.
- $$\begin{aligned} &= -\frac{u^{3}}{3} + \frac{u^{5}}{5} + C \\ &= -\frac{\cos^{3}x}{3} + \frac{\cos^{5}x}{5} + C. \end{aligned}$$
Answer: $-\frac{\cos^{3}x}{3} + \frac{\cos^{5}x}{5} + C$
Worked example: An even power
Find $\int \sin^{2}x\,dx$, then $\int_0^{\pi}\sin^{2}x\,dx$.
- $\int \frac{1 - \cos 2x}{2}\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$.
- From 0 to $\pi$: $\frac{\pi}{2} - 0 - (0 - 0) = \frac{\pi}{2}$.
Answer: $\frac{x}{2} - \frac{\sin 2x}{4} + C$; the definite integral is $\frac{\pi}{2}$.
Worked example: Tangent and secant
Find $\int \tan^{3}x\sec x\,dx$.
- The power of $\tan x$ is odd: save $\sec x\tan x\,dx$, which is $d(\sec x)$.
- What is left is $\tan^{2}x = \sec^{2}x - 1$. With $u = \sec x$: $\int (u^{2} - 1)\,du = \frac{u^{3}}{3} - u + C$.
- $= \frac{\sec^{3}x}{3} - \sec x + C$.
Answer: $\frac{\sec^{3}x}{3} - \sec x + C$
Common mistake
Substituting $u = \cos x$ when no spare $\sin x$ is available, as in $\int \cos^{2}x\,dx$. Then $du = -\sin x\,dx$ has nothing to match, and you end up with leftover $x$'s next to $u$'s. Check after substituting: if any $x$ is left in the integrand, the substitution did not work. For even powers, reduce the power first.
Products with different angles, like $\int \sin 3x\cos x\,dx$, use the product-to-sum identity $$\begin{aligned} \sin A\cos B &= \tfrac{1}{2}\sin(A + B) \\ &\quad + \tfrac{1}{2}\sin(A - B), \end{aligned}$$ which turns the product into two easy terms.