Integrating Powers of Sine & Cosine
What happens when you need to integrate $\sin^3 x$, $\cos^4 x$, or products like $\sin^2 x\cos^3 x$? You can't just "reverse the chain rule." The trick is a two-move strategy: split off one factor and use a Pythagorean identity to convert everything into a single substitutable form. Master the odd-power trick and the double-angle method and you'll handle any power of sine or cosine the HSC can throw at you.
You already know $\displaystyle\int \sin x\,dx = -\cos x + C$ and $\displaystyle\int \sin^2 x\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$ (from Lesson 6). Without using any new formula — can you think of a way to handle $\displaystyle\int \sin^3 x\,dx$? Write your instinct below.
Every integral of a pure power of sine or cosine uses one of two strategies, depending on whether the power is odd or even:
Strategy A — Odd power: Split off one factor and convert the remainder using $\sin^2 x = 1 - \cos^2 x$ (or $\cos^2 x = 1 - \sin^2 x$), then substitute $u = \cos x$ (or $u = \sin x$).
Strategy B — Even power: Use the double-angle identity to reduce the power: $\sin^2 x = \dfrac{1-\cos 2x}{2}$, $\cos^2 x = \dfrac{1+\cos 2x}{2}$.
Key facts
- $\sin^2 x = 1 - \cos^2 x$ (Pythagorean) and $\sin^2 x = \dfrac{1-\cos 2x}{2}$ (double-angle)
- For odd $n$: split off one factor, convert, substitute $u = \cos x$
- For even $n$: use the half-angle form to reduce the power
Concepts
- Why $u = \cos x \Rightarrow du = -\sin x\,dx$ makes the integral tractable for odd sine powers
- How double-angle identities convert hard integrals into standard ones
- When to split a mixed product $\sin^m x \cos^n x$
Skills
- Integrate $\sin^3 x$, $\cos^3 x$, $\sin^4 x$, $\cos^4 x$ and products
- Correctly change limits in definite integrals after substitution
- Check an antiderivative by differentiation
To integrate $\sin^n x$ when $n$ is odd, rewrite the integrand by splitting off one $\sin x$ and converting the rest via $\sin^2 x = 1 - \cos^2 x$:
Now let $u = \cos x$, so $du = -\sin x\,dx$, i.e. $\sin x\,dx = -du$. The integral becomes a polynomial in $u$, which you can integrate term-by-term, then substitute back.
Example — $\displaystyle\int \sin^3 x\,dx$:
Check: Differentiate: $\dfrac{d}{dx}\!\left(-\cos x + \tfrac{\cos^3 x}{3}\right) = \sin x - \cos^2 x \sin x = \sin x(1-\cos^2 x) = \sin^3 x$ ✓
To integrate $\sin^n x$ when $n$ is odd , rewrite the integrand by splitting off one $\sin x$ and converting the rest via $\sin^2 x = 1 - \cos^2 x$:
Pause — copy Strategy A: $\int\sin^n x\,dx$ for odd $n$ — split off $\sin x$, replace $\sin^{n-1}x$ using $1-\cos^2 x$, substitute $u=\cos x$ into your book.
Quick check: To evaluate $\displaystyle\int \cos^3 x\,dx$, which substitution is most appropriate after rewriting $\cos^3 x = (1-\sin^2 x)\cos x$?
We just saw Strategy A for odd $n$: split $\sin^n x=\sin^{n-1}x\cdot\sin x$, use $\sin^2 x=1-\cos^2 x$, then substitute $u=\cos x$ so $du=-\sin x\,dx$. That raises a question: when $n$ is even, the substitution approach doesn't simplify things — which identity do you apply repeatedly, and how far do you have to reduce before you can integrate? This card answers it → for even $n$, apply $\sin^2 x=\frac{1-\cos2x}{2}$ repeatedly until all powers are 1.
When $n$ is even, substitution won't simplify things. Instead, use the half-angle (power-reduction) forms:
Apply these once to halve the power. If the resulting integrand still has an even power (e.g. $\cos^4 x$ gives $\cos^2 2x$), apply the identity again.
Example — $\displaystyle\int \sin^4 x\,dx$:
Now integrate term-by-term:
When $n$ is even , substitution won't simplify things. Instead, use the half-angle (power-reduction) forms:
Pause — copy Strategy B: $\int\sin^n x\,dx$ for even $n$ — apply $\sin^2 x=\frac{1-\cos2x}{2}$ repeatedly until only $\cos(2kx)$ terms with power 1 remain into your book.
Did you get this? True or false: $\displaystyle\int \cos 4x\,dx = \dfrac{\sin 4x}{4} + C$.
Worked examples · 3 in a row, reveal as you go
Find $\displaystyle\int \sin^5 x\,dx$.
Find $\displaystyle\int \cos^4 x\,dx$.
Evaluate $\displaystyle\int_0^{\pi/2} \sin^2 x\cos^3 x\,dx$.
$\displaystyle\int_0^1 u^2(1-u^2)\,du = \int_0^1 (u^2-u^4)\,du$
Fill the gap: When finding $\displaystyle\int \sin^3 x\,dx$, we let $u = \cos x$ so that $du = $ (include the $dx$).
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the correct approach for $\displaystyle\int \sin^4 x\,dx$ is to use the double-angle identity $\sin^2 x = \dfrac{1-\cos 2x}{2}$, not a $u$-substitution directly.
Activities · practice with the ideas
Find $\displaystyle\int \cos^3 x\,dx$. (Hint: $\cos^3 x = (1-\sin^2 x)\cos x$.)
Use the double-angle identity to find $\displaystyle\int \cos^2 x\,dx$.
Evaluate $\displaystyle\int_0^{\pi/2} \cos^3 x\,dx$ exactly.
Find $\displaystyle\int \sin^3 x\cos^2 x\,dx$. (Identify which power is odd.)
Verify that $\dfrac{d}{dx}\!\left(-\cos x + \dfrac{\cos^3 x}{3}\right) = \sin^3 x$. Show all steps.
Odd one out: Three of these are valid steps in an integration of a power of sine/cosine. Which one is NOT a valid step?
At the start you tried to guess what $\displaystyle\int \sin^3 x\,dx$ could equal. The answer is $-\cos x + \dfrac{\cos^3 x}{3} + C$.
Could you have guessed it? The key insight is that no simple "reverse rule" exists — you must use the Pythagorean identity to manufacture a substitution. Once you see the $\sin x$ as a $-du$ factor, the whole integral collapses into a polynomial.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find $\displaystyle\int \cos^3 x\,dx$, showing your substitution clearly. (2 marks)
Q2. Evaluate $\displaystyle\int_0^{\pi/2} \sin^3 x\,dx$ exactly, showing the change of limits. (3 marks)
Q3. Find $\displaystyle\int \sin^2 x\cos^3 x\,dx$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $\sin x - \tfrac{\sin^3 x}{3} + C$ · 2. $\tfrac{x}{2}+\tfrac{\sin 2x}{4}+C$ · 3. $\tfrac{2}{3}$ · 4. $-\tfrac{\cos^3 x}{3}+\tfrac{\cos^5 x}{5}+C$ · 5. chain rule gives $\sin x - \sin x\cos^2 x = \sin x(1-\cos^2 x) = \sin^3 x$ ✓
Q1 (2 marks): $\cos^3 x = (1-\sin^2 x)\cos x$. Let $u=\sin x$, $du=\cos x\,dx$ [1]. $\int(1-u^2)\,du = u - \tfrac{u^3}{3}+C = \sin x - \tfrac{\sin^3 x}{3}+C$ [1].
Q2 (3 marks): $\sin^3 x = (1-\cos^2 x)\sin x$. Let $u=\cos x$, $du=-\sin x\,dx$ [1]. Limits: $x=0\to u=1$; $x=\tfrac{\pi}{2}\to u=0$ [1]. $\int_1^0(1-u^2)(-du)=\int_0^1(1-u^2)\,du=\left[u-\tfrac{u^3}{3}\right]_0^1 = 1-\tfrac{1}{3}=\tfrac{2}{3}$ [1].
Q3 (3 marks): cos power odd → split: $\sin^2 x\cos^3 x = \sin^2 x(1-\sin^2 x)\cos x$ [1]. Let $u=\sin x$, $du=\cos x\,dx$ [1]. $\int u^2(1-u^2)\,du = \tfrac{u^3}{3}-\tfrac{u^5}{5}+C = \tfrac{\sin^3 x}{3}-\tfrac{\sin^5 x}{5}+C$ [1].
Five timed questions on integrating powers of sine and cosine. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering trig integration questions. Lighter alternative to the boss.
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