Integrating Trigonometric Powers

Look for a derivative pair

For a product of sine and cosine powers, the exponents suggest a useful substitution. An odd sine power can supply a factor $\sin x\,dx$ for $u=\cos x$ . An odd cosine power can supply $\cos x\,dx$ for $u=\sin x$ . Rewrite the remaining even power with $\sin^2x+\cos^2x=1$ .

When both powers are even, half-angle identities reduce the exponents:

\sin^2x=\frac{1-\cos2x}{2},

\cos^2x=\frac{1+\cos2x}{2}.

These transformations come from the double-angle formulas. They change the integrand algebraically; no approximation is involved.

Visual guide

VISUAL GUIDEUse symmetry and a double-angle identity

sin²x
Mean level 1/2

The identity sin²x = (1 − cos 2x)/2 shows a nonnegative wave centered at 1/2. Over [0, π], the oscillating part cancels, leaving area π/2.

Worked example: an odd power

For $\int\sin^3x\cos^2x\,dx$ , retain one sine factor and replace the other two:

\sin^3x\cos^2x=\sin x(1-\cos^2x)\cos^2x.

Let $u=\cos x$ , so $du=-\sin x\,dx$ . The integral becomes $-\int(u^2-u^4)du=-u^3/3+u^5/5+C$ . Returning to $x$ gives $-\cos^3x/3+\cos^5x/5+C$ . The negative sign originates in the cosine derivative.

Worked example: two even powers

Since $\sin^2x\cos^2x=\tfrac14\sin^2(2x)=\tfrac18(1-\cos4x)$ ,

\int\sin^2x\cos^2x\,dx=\frac{x}{8}-\frac{\sin4x}{32}+C.

The frequency $4$ produces the extra denominator when integrating cosine. Missing that factor is easy to detect by differentiation.

Tangent and secant patterns

For $\tan^m x\sec^n x$ , an even positive secant power can provide $\sec^2x\,dx$ for $u=\tan x$ , using $\sec^2x=1+\tan^2x$ . An odd positive tangent power together with at least one secant factor can provide $\sec x\tan x\,dx$ for $u=\sec x$ .

For example, $\int\tan^2x\sec^2x\,dx=\tan^3x/3+C$ . These patterns do not cover every combination; $\int\sec x\,dx=\ln|\sec x+\tan x|+C$ is a useful separate identity, and some cases need integration by parts. All formulas are used on intervals avoiding poles.

Practice

Integrate $\sin x\cos^4x$ .
Evaluate $\int_0^\pi\sin^2x\,dx$ .
Integrate $\tan^3x\sec^2x$ .

Show worked solutions

$u=\cos x$ gives $-\cos^5x/5+C$ .
The half-angle identity gives $[x/2-\sin2x/4]_0^\pi=\pi/2$ .
$u=\tan x$ gives $\tan^4x/4+C$ on any interval where tangent is defined.