Powers of Sine and Cosine
\(\int \sin^m x \cos^n x\,dx\) looks like it needs a table of tricks, but one question settles it: is either power odd? If so, the Pythagorean identity sets up a clean u-substitution; if both powers are even, the half-angle identities take over.
By the end you'll be able to classify \(\int \sin^m x \cos^n x\,dx\) by the parity of its powers, reserve the right factor for \(du\) when a power is odd, and switch to the power-reducing identities when both powers are even.
Predict: if the sine power is odd, which factor do you save for \(du\) — a \(\sin x\) or a \(\cos x\)? Set m to an odd value below and check which chip turns orange.
Each chip is one factor of the integrand \(\sin^m x \cos^n x\,dx\). Slide m and n to watch the strategy switch: an odd power donates one saved factor to \(du\), the leftover even power is converted by the Pythagorean identity, and the other function becomes u. Make both powers even and no chip can be saved — the dashed pairs switch to half-angle rewrites. Hover or tab to any chip for its exact role.
Odd sine power → u = cos x, du = −sin x dx · rewritten: −∫(1 − u²)u2 du
Integrals of products of powers of sine and cosine are evaluated by using the Pythagorean identity sin²+cos²=1 to reserve one factor for a u-substitution when a power is odd, or the power-reducing (half-angle) identities when both powers are even.
Why does an odd power unlock a substitution? Because \(\frac{d}{dx}\sin x = \cos x\) and \(\frac{d}{dx}\cos x = -\sin x\): each function's derivative is (up to sign) the other function. Say the cosine power \(n\) is odd. Peel off a single \(\cos x\) to serve as \(du\) for \(u = \sin x\); what remains is \(\cos^{n-1} x\) with an even exponent, and even powers convert exactly: \(\cos^{n-1} x = (1 - \sin^2 x)^{(n-1)/2} = (1-u^2)^{(n-1)/2}\), a genuine polynomial in \(u\). The whole integrand collapses to a polynomial times \(du\). If instead the sine power is odd, the mirror move works: save one \(\sin x\), substitute \(u = \cos x\), and remember that \(du = -\sin x\,dx\) carries a minus sign. When both powers are even, no single factor can be spared — whichever you peel off leaves an odd power behind, and \((1-u^2)\) raised to a half-integer is no polynomial. That's when the power-reducing identities take over, trading every squared factor for a first power of \(\cos 2x\).
In practice, run a two-step checklist. Step 1 — check parity. Odd cosine power → save one \(\cos x\), set \(u = \sin x\), \(du = \cos x\,dx\). Odd sine power → save one \(\sin x\), set \(u = \cos x\), \(du = -\sin x\,dx\). Both even → apply \(\sin^2 x = \frac{1-\cos 2x}{2}\) and \(\cos^2 x = \frac{1+\cos 2x}{2}\), expand, and integrate term by term (re-applying the identities if a \(\cos^2 2x\) appears). Step 2 — check the integrand shape first. This strategy is for pure products of sine and cosine powers; if an algebraic factor like \(x\) is riding along, as in \(\int x\cos x\,dx\), no identity removes it — that's a job for integration by parts from the previous lesson, not for this playbook.
Evaluate \(\int \sin^3 x \cos^2 x\,dx\). The sine power (3) is odd, so save one \(\sin x\) and convert the rest: \(\sin^3 x = \sin^2 x \cdot \sin x = (1 - \cos^2 x)\sin x\). Then \(\int \sin^3 x \cos^2 x\,dx = \int (1-\cos^2 x)\cos^2 x \sin x\,dx\). Let \(u = \cos x\), \(du = -\sin x\,dx\), so \(\sin x\,dx = -du\): \(-\int(1-u^2)u^2\,du = -\int (u^2 - u^4)\,du = -\dfrac{u^3}{3} + \dfrac{u^5}{5} + C\). Back-substitute \(u = \cos x\): \(-\dfrac{\cos^3 x}{3} + \dfrac{\cos^5 x}{5} + C\).
Evaluate \(\int \cos^3 x\,dx\). The cosine power (3) is odd, so save one \(\cos x\) for \(du\) and convert the rest: \(\cos^3 x = (1 - \sin^2 x)\cos x\). Let \(u = \sin x\), so \(du = \cos x\,dx\) — no minus sign this time. Now finish it: \(\int \cos^3 x\,dx = \int (1 - u^2)\,du = \) ____
Reveal the answer
\(\int (1-u^2)\,du = u - \dfrac{u^3}{3} + C = \sin x - \dfrac{\sin^3 x}{3} + C\). Set m = 0 and n = 3 on the sliders above and confirm the widget saves one cosine chip and turns the remaining pair into \(1 - u^2\).
More info — why the both-even case genuinely needs the half-angle identities
Try to force a substitution on \(\int \cos^4 x\,dx\). Saving one \(\cos x\) for \(du = \cos x\,dx\) leaves \(\cos^3 x\) — an odd power — and writing it in terms of \(u = \sin x\) gives \((1-u^2)^{3/2}\), a square root, not a polynomial. The substitution didn't simplify anything; it just relocated the difficulty. The half-angle identities attack from a different direction: instead of changing variables, they lower the degree, turning \(\cos^4 x = \left(\frac{1+\cos 2x}{2}\right)^2\) into a sum of constants, \(\cos 2x\), and \(\cos^2 2x\) — the last of which yields to one more application of the same identity. The strategy table in the OpenStax chapter under Dive deeper lays out all the cases side by side and is worth a read before drilling.
Check your understanding
Evaluate \(\int \sin x \cos^4 x\,dx\).
Why can't \(\int \sin^2 x \cos^4 x\,dx\) be evaluated by substituting \(u = \sin x\) (or \(u = \cos x\))?
For \(\int \sin^4 x \cos^3 x\,dx\), which factor do you reserve, and what substitution follows?
You face two integrals: \(\int x \cos x\,dx\) and \(\int \sin^3 x \cos x\,dx\). Which technique fits which?
Recap
- For \(\int \sin^m x \cos^n x\,dx\), only the parity of the powers matters: check odd vs. even before anything else.
- Odd cosine power: save one \(\cos x\), substitute \(u = \sin x\), and convert the even leftover with \(\cos^2 x = 1 - \sin^2 x\).
- Odd sine power: save one \(\sin x\), substitute \(u = \cos x\) — and keep the minus sign from \(du = -\sin x\,dx\).
- Both powers even: no factor can be spared for \(du\); use the power-reducing identities \(\sin^2 x = \frac{1-\cos 2x}{2}\), \(\cos^2 x = \frac{1+\cos 2x}{2}\) and integrate term by term.
Dive deeper
- Paul's Online Math Notes — Integrals Involving Trig Functions Drill the odd-power and both-even-power cases across worked examples.
- OpenStax Calculus Volume 2 — 3.2 Trigonometric Integrals Read the general strategy table for ∫cos^j x sin^k x dx before the checkpoints.
Sources
- Integrals of Powers of Sine and Cosine