3.2 Trigonometric Integrals

Big idea

Powers of sine and cosine do not yield to any of the rules you already have. There is no product rule for antiderivatives that turns $\sin^5 x$ into something familiar, and substitution alone gets stuck because $\sin x$ and its derivative $\cos x$ are not both present. The fix is to rewrite the integrand first, using the Pythagorean identity, so that a substitution becomes available.

The whole section runs on one observation. If you can peel off a single factor of $\sin x$, then that factor is exactly what $u = \cos x$ needs for its $du$, and the identity $\sin^2 x = 1 - \cos^2 x$ converts everything that remains into a polynomial in $\cos x$. So the question for any product of powers of sine and cosine is whether some power is odd. An odd power supplies the spare factor; an even power does not.

When both powers are even, no factor can be spared without leaving a square root behind, and the strategy changes completely. The half-angle identities trade a squared trigonometric function for a constant plus a cosine of double the angle. The powers drop, the angle doubles, and repeating the move eventually reduces the integrand to terms you can integrate directly.

Secant and tangent behave the same way with different bookkeeping, because their derivatives feed each other: $\frac{d}{dx}\tan x = \sec^2 x$ and $\frac{d}{dx}\sec x = \sec x\tan x$. The identity that links them is $\sec^2 x = 1 + \tan^2 x$. The parity you look for is different, but the plan is identical: reserve the derivative factor, convert what remains with an identity, substitute.

Decoder

A product of powers of sine and cosine becomes a polynomial substitution whenever one of the two powers is odd; when both are even, the half-angle identities must lower the powers first.

The word carrying the content is odd. Splitting $\sin^{2k+1}x$ as $(\sin^2 x)^k \sin x$ and replacing $\sin^2 x$ by $1 - \cos^2 x$ leaves $(1 - \cos^2 x)^k \sin x\,dx$, and with $u = \cos x$ you have $du = -\sin x\,dx$ waiting at the end. Nothing is left over. The same argument with the roles swapped handles an odd power of cosine using $u = \sin x$.

If both powers are odd, either choice works, and it is worth taking the smaller odd power as the one you split, because that leaves the lower-degree polynomial to expand.

The classic mistake is trying to substitute before rewriting. Setting $u = \sin x$ in $\int \sin^4 x\,dx$ produces $\int u^4\,\frac{du}{\cos x}$, and the leftover $\cos x$ cannot be expressed in $u$ without a square root and a sign ambiguity. Rewrite first, substitute second, and the leftover never appears.

Definitions and results

The Pythagorean identities. The three forms used throughout are

$$ \sin^2 x + \cos^2 x = 1, \qquad \sec^2 x = 1 + \tan^2 x, \qquad \csc^2 x = 1 + \cot^2 x $$

Odd power of sine. If the power of $\sin x$ is odd, save one factor of $\sin x$, convert the rest with $\sin^2 x = 1 - \cos^2 x$, and substitute $u = \cos x$, $du = -\sin x\,dx$.

Odd power of cosine. If the power of $\cos x$ is odd, save one factor of $\cos x$, convert the rest with $\cos^2 x = 1 - \sin^2 x$, and substitute $u = \sin x$, $du = \cos x\,dx$.

Both powers even. Use the half-angle identities

$$ \sin^2 x = \frac{1 - \cos 2x}{2}, \qquad \cos^2 x = \frac{1 + \cos 2x}{2} $$

and the double-angle identity $\sin x\cos x = \frac{1}{2}\sin 2x$. Each application halves the exponent and doubles the angle; apply it as many times as needed.

Odd power of tangent with secant present. Save a factor of $\sec x\tan x$, convert the remaining even power of tangent with $\tan^2 x = \sec^2 x - 1$, and substitute $u = \sec x$.

Even power of secant with tangent present. Save a factor of $\sec^2 x$, convert the remaining even power of secant with $\sec^2 x = 1 + \tan^2 x$, and substitute $u = \tan x$.

The two base antiderivatives. These are worth memorizing because every secant-tangent problem eventually lands on one of them:

$$ \int \tan x\,dx = -\ln|\cos x| + C, \qquad \int \sec x\,dx = \ln\big|\sec x + \tan x\big| + C $$

Different angles inside. For a product of sines and cosines with unequal arguments, use the product-to-sum identities, for instance

$$ \sin A\cos B = \tfrac{1}{2}\big[\sin(A+B) + \sin(A-B)\big] $$

which converts a product into a sum of single trigonometric terms you can integrate one at a time.

Worked examples

An odd power of sine

Evaluate $\int \sin^3 x\cos^2 x\,dx$.

The power of sine is odd, so split off one factor and convert the rest: $\sin^3 x = (1-\cos^2 x)\sin x$. With $u = \cos x$ and $du = -\sin x\,dx$,

$$ \int (1-\cos^2 x)\cos^2 x\,\sin x\,dx = -\int (1-u^2)u^2\,du = -\int \big(u^2 - u^4\big)\,du $$

That is $-\frac{u^3}{3} + \frac{u^5}{5}$, so

$$ \int \sin^3 x\cos^2 x\,dx = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C $$

Check by differentiating: $\cos^4 x(-\sin x) - \cos^2 x(-\sin x) = \sin x\cos^2 x(1 - \cos^2 x) = \sin^3 x\cos^2 x$.

Both powers even

Evaluate $\int \sin^2 x\cos^2 x\,dx$.

No factor can be spared. The fastest route uses the double-angle identity on the product itself: $\sin x\cos x = \frac{1}{2}\sin 2x$, so $\sin^2 x\cos^2 x = \frac{1}{4}\sin^2 2x$. Now apply the half-angle identity to $\sin^2 2x$:

$$ \frac{1}{4}\sin^2 2x = \frac{1}{4}\cdot\frac{1 - \cos 4x}{2} = \frac{1}{8} - \frac{\cos 4x}{8} $$

Integrating term by term,

$$ \int \sin^2 x\cos^2 x\,dx = \frac{x}{8} - \frac{\sin 4x}{32} + C $$

The average value of the integrand over a full period should be $\frac18$, since $\sin^2$ and $\cos^2$ each average $\frac12$ and their product averages $\frac18$. The linear term matches that, and the oscillating term contributes nothing over a period.

Tangent and secant together

Evaluate $\int \tan^3 x\sec^3 x\,dx$.

The power of tangent is odd and a secant is available, so reserve $\sec x\tan x\,dx$ and convert the leftover $\tan^2 x$ into $\sec^2 x - 1$:

$$ \int \tan^3 x\sec^3 x\,dx = \int \big(\sec^2 x - 1\big)\sec^2 x\,\big(\sec x\tan x\,dx\big) $$

With $u = \sec x$ the integral is $\int (u^2-1)u^2\,du = \frac{u^5}{5} - \frac{u^3}{3}$, so

$$ \int \tan^3 x\sec^3 x\,dx = \frac{\sec^5 x}{5} - \frac{\sec^3 x}{3} + C $$

Unequal angles

Evaluate $\int \sin 3x\cos 5x\,dx$.

The arguments differ, so no Pythagorean identity applies. Use the product-to-sum identity with $A = 3x$ and $B = 5x$:

$$ \sin 3x\cos 5x = \tfrac{1}{2}\big[\sin 8x + \sin(-2x)\big] = \tfrac{1}{2}\big[\sin 8x - \sin 2x\big] $$

Each piece integrates on sight:

$$ \int \sin 3x\cos 5x\,dx = -\frac{\cos 8x}{16} + \frac{\cos 2x}{4} + C $$

Differentiating gives $\frac{8\sin 8x}{16} - \frac{2\sin 2x}{4} = \frac{1}{2}(\sin 8x - \sin 2x)$, which is the rewritten integrand.

Practice

Start with the antiderivatives themselves. These problems ask for integrals of trigonometric and exponential expressions where the identity work is the main step and the final antiderivative is standard.

Then drill the substitution that every one of these problems ends with. Identifying $u$ and matching $du$ against the factor you reserved is the mechanical half of the technique, and it should be fast.

Quiz

Five items on odd and even powers of sine and cosine, secant-tangent combinations, and products with different angles.