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Module 8 · L6 of 20 ~35 min ⚡ +95 XP available

Integrating $\sin^2 nx$ and $\cos^2 nx$

You cannot integrate $\sin^2 x$ directly using the power rule. But a single double-angle identity rewrites it as a sum of simple terms that integrate instantly. Once you know the two key identities, integrating any $\sin^2 nx$ or $\cos^2 nx$ — including over definite intervals — becomes a two-step routine.

Today's hook — Without any calculation, what do you think $\displaystyle\int_0^\pi \sin^2 x\,dx$ equals? Is it more than, less than, or equal to $\pi/2$? Jot your guess and reasoning. You'll find out after card 06.

0/5QUESTS

Recall — your gut answer first

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Write down the double-angle identity for $\cos 2\theta$ in terms of $\sin^2 \theta$ and then in terms of $\cos^2 \theta$. From memory — don't look it up yet.

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The two identities you need

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The double-angle formula $\cos 2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$ can be rearranged to give the two key identities for integration:

Rearranging each form of the double-angle identity:

From $\cos 2\theta = 1 - 2\sin^2\theta$: solve for $\sin^2\theta$
From $\cos 2\theta = 2\cos^2\theta - 1$: solve for $\cos^2\theta$

$\sin^2\theta = \dfrac{1-\cos 2\theta}{2}$ $\cos^2\theta = \dfrac{1+\cos 2\theta}{2}$

Memory hook

$\sin^2$ has a minus sign: $\frac{1-\cos 2\theta}{2}$. $\cos^2$ has a plus sign: $\frac{1+\cos 2\theta}{2}$. The letter that comes earlier in "s-c" has the minus.

Generalise with $n$

Replace $\theta$ with $nx$: $\sin^2 nx = \frac{1-\cos 2nx}{2}$ and $\cos^2 nx = \frac{1+\cos 2nx}{2}$. The frequency doubles from $n$ to $2n$.

Sum to 1

$\sin^2\theta + \cos^2\theta = 1$ always. Adding the two identities: $\frac{1-\cos 2\theta}{2} + \frac{1+\cos 2\theta}{2} = 1$ — a useful check.

What you'll master

Know

Key facts

$\sin^2 nx = \dfrac{1-\cos 2nx}{2}$ and $\cos^2 nx = \dfrac{1+\cos 2nx}{2}$
$\displaystyle\int \sin^2 nx\,dx = \frac{x}{2} - \frac{\sin 2nx}{4n} + C$
$\displaystyle\int \cos^2 nx\,dx = \frac{x}{2} + \frac{\sin 2nx}{4n} + C$

Understand

Concepts

Why the double-angle identity is needed (power rule does not apply to $\sin^2 x$)
How the frequency doubles when the identity is applied ($n \to 2n$)
Why definite integrals of $\sin^2$ and $\cos^2$ over a full period both equal half the period length

Can do

Skills

Apply the double-angle identity to rewrite $\sin^2 nx$ or $\cos^2 nx$ before integrating
Evaluate indefinite and definite integrals involving $\sin^2 nx$ and $\cos^2 nx$
Find the exact area under a squared-trig curve over a specified interval

Key terms

Double-angle identityA trigonometric identity relating a function of $2\theta$ to functions of $\theta$. The key forms are: $\cos 2\theta = 1 - 2\sin^2\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$.

$\sin^2 nx$The square of $\sin(nx)$. Cannot be integrated by the power rule. Requires the identity $\sin^2 nx = \frac{1}{2}(1-\cos 2nx)$ first.

$\cos^2 nx$The square of $\cos(nx)$. Requires the identity $\cos^2 nx = \frac{1}{2}(1+\cos 2nx)$ before integrating.

Period halving$\sin^2 x$ has period $\pi$ (half the period of $\sin x$). After applying the identity, the cosine term has frequency $2n$.

$\int \cos(kx)\,dx$$= \dfrac{\sin(kx)}{k} + C$. This is the result needed for the cosine term after applying the double-angle identity.

Symmetry resultOver a full period: $\int_0^\pi \sin^2 x\,dx = \int_0^\pi \cos^2 x\,dx = \dfrac{\pi}{2}$, since $\sin^2 + \cos^2 = 1$ and both have equal area by symmetry.

Integrating $\sin^2 nx$

core concept

The power rule $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$ does not apply to $\int \sin^2 x\,dx$ — we cannot "integrate the function and add 1 to the power". Instead, apply the double-angle identity first:

$$\sin^2 nx = \frac{1-\cos 2nx}{2}$$

Then integrate term by term:

$$\int \sin^2 nx\,dx = \int \frac{1-\cos 2nx}{2}\,dx = \frac{x}{2} - \frac{\sin 2nx}{4n} + C$$

Where does $4n$ come from? $\int \cos 2nx\,dx = \dfrac{\sin 2nx}{2n}$. Dividing by 2 (from the $\frac{1}{2}$ outside): gives $\dfrac{\sin 2nx}{4n}$.

Special case $n = 1$. $\displaystyle\int \sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$. For the definite integral $\int_0^\pi \sin^2 x\,dx$: at $x=\pi$, $\sin 2\pi = 0$; at $x=0$, $\sin 0 = 0$. Result: $\frac{\pi}{2} - 0 = \frac{\pi}{2}$.

The power rule $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$ does not apply to $\int \sin^2 x\,dx$ — we cannot "integrate the function and add 1 to the power". Instead, apply the double-angle identity first:

Pause — copy the identity $\sin^2(nx)=\frac{1-\cos(2nx)}{2}$ and the result $\int\sin^2(nx)\,dx=\frac{x}{2}-\frac{\sin(2nx)}{4n}+C$ into your book.

Quick check: Which identity is used to integrate $\sin^2 3x$?

Integrating $\cos^2 nx$

core concept

We just saw that $\int\sin^2(nx)\,dx$ uses the power-reduction identity $\sin^2(nx)=\frac{1-\cos(2nx)}{2}$, giving $\frac{x}{2}-\frac{\sin(2nx)}{4n}+C$. That raises a question: for $\int\cos^2(nx)\,dx$, the analogous identity has a plus sign — does the integration procedure differ, and what is the result? This card answers it → use $\cos^2(nx)=\frac{1+\cos(2nx)}{2}$ to get $\frac{x}{2}+\frac{\sin(2nx)}{4n}+C$; only the sign of the second term changes.

The method is identical, but the identity has a plus sign:

$$\cos^2 nx = \frac{1+\cos 2nx}{2}$$

Integrate term by term:

$$\int \cos^2 nx\,dx = \frac{x}{2} + \frac{\sin 2nx}{4n} + C$$

Revisit the hook. $\displaystyle\int_0^\pi \sin^2 x\,dx = \frac{\pi}{2}$. This equals half the length of the interval $[0,\pi]$. Notice $\int_0^\pi \cos^2 x\,dx$ also equals $\dfrac{\pi}{2}$ by the same calculation (the $\sin 2x$ terms vanish). Since $\sin^2 x + \cos^2 x = 1$, both areas must sum to $\pi$, so each is $\dfrac{\pi}{2}$.

The method is identical, but the identity has a plus sign:

Pause — copy the identity $\cos^2(nx)=\frac{1+\cos(2nx)}{2}$ and the result $\int\cos^2(nx)\,dx=\frac{x}{2}+\frac{\sin(2nx)}{4n}+C$ into your book.

Did you get this? True or false: $\displaystyle\int \cos^2 2x\,dx = \frac{x}{2} + \frac{\sin 4x}{8} + C$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · INDEFINITE INTEGRAL

Find $\displaystyle\int \sin^2 3x\,dx$.

Apply the identity: $\sin^2 3x = \dfrac{1-\cos 6x}{2}$.

Here $n = 3$, so $2n = 6$. The identity replaces $\sin^2 3x$ with a form that can be integrated directly.

PROBLEM 2 · DEFINITE INTEGRAL

Evaluate $\displaystyle\int_0^{\pi/2} \cos^2 x\,dx$.

$\cos^2 x = \dfrac{1+\cos 2x}{2}$, so $\displaystyle\int_0^{\pi/2}\cos^2 x\,dx = \int_0^{\pi/2}\frac{1+\cos 2x}{2}\,dx$.

Apply the identity with $n=1$ before substituting limits.

PROBLEM 3 · AREA PROBLEM

Find the exact area enclosed between $y = \cos^2 x$ and $y = \sin^2 x$ for $0 \leq x \leq \dfrac{\pi}{4}$.

For $0 \leq x \leq \dfrac{\pi}{4}$, $\cos x \geq \sin x \geq 0$, so $\cos^2 x \geq \sin^2 x$. Area $= \displaystyle\int_0^{\pi/4}(\cos^2 x - \sin^2 x)\,dx$.

The upper curve is $y = \cos^2 x$ on this interval. (At $x = \pi/4$, both equal $\frac{1}{2}$.) The integrand is the difference of the two curves.

Fill the gap: $\displaystyle\int \cos^2 x\,dx = \dfrac{x}{2} + $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Using the wrong sign

$\sin^2 nx = \dfrac{1-\cos 2nx}{2}$ has a minus; $\cos^2 nx = \dfrac{1+\cos 2nx}{2}$ has a plus. Swapping these signs is the single most common error in trig integration. If you confuse them, every integral result will be wrong. Use the memory hook: $\sin$ comes before $\cos$ alphabetically, and $\sin$ gets the minus.

Trap 02

Forgetting to divide by $2n$ after integrating $\cos 2nx$

$\displaystyle\int \cos 2nx\,dx = \dfrac{\sin 2nx}{2n}$, not $\sin 2nx$. After dividing by the outer $\frac{1}{2}$ from the identity, the denominator becomes $4n$. A common error is to write $\frac{\sin 2nx}{2}$ instead of $\frac{\sin 2nx}{4n}$.

Trap 03

Applying the power rule to $\sin^2 x$

Writing $\int \sin^2 x\,dx = \dfrac{\sin^3 x}{3} + C$ is incorrect — the power rule only applies to $\int x^n\,dx$, not to $\int (\sin x)^n\,dx$. Always apply the double-angle identity first before integrating any squared trig function.

Did you get this? True or false: $\displaystyle\int \sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Find $\displaystyle\int \sin^2 2x\,dx$.

Find $\displaystyle\int \cos^2 3x\,dx$.

Evaluate $\displaystyle\int_0^{\pi} \sin^2 2x\,dx$.

Find the area under $y = \cos^2 x$ from $x = 0$ to $x = \pi$.

Using the identity $\cos^2 x - \sin^2 x = \cos 2x$, evaluate $\displaystyle\int_0^{\pi/4}(\cos^2 x - \sin^2 x)\,dx$.

Odd one out: Three of these results are correct. Which one is WRONG?

Revisit your thinking

At the start you estimated $\displaystyle\int_0^\pi \sin^2 x\,dx$. The exact answer is $\dfrac{\pi}{2} \approx 1.57$.

Since $0 \leq \sin^2 x \leq 1$ on $[0,\pi]$ and the curve is symmetric about $x = \pi/2$, the area is exactly half the rectangle of area $\pi \times 1 = \pi$. This makes geometric sense.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Find $\displaystyle\int \sin^2 4x\,dx$. (2 marks)

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ApplyBand 43 marks

Q2. Evaluate $\displaystyle\int_0^{\pi/6} \cos^2 3x\,dx$. (3 marks)

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AnalyseBand 53 marks

Q3. Find the exact area enclosed between $y = \sin^2 x$ and $y = \cos^2 x$ for $0 \leq x \leq \dfrac{\pi}{2}$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\sin^2 2x = \frac{1-\cos 4x}{2}$. $\int \sin^2 2x\,dx = \frac{x}{2} - \frac{\sin 4x}{8} + C$.

2. $\cos^2 3x = \frac{1+\cos 6x}{2}$. $\int \cos^2 3x\,dx = \frac{x}{2} + \frac{\sin 6x}{12} + C$.

3. $\int_0^{\pi}\sin^2 2x\,dx = \left[\frac{x}{2}-\frac{\sin 4x}{8}\right]_0^{\pi} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

4. Area $= \int_0^{\pi}\cos^2 x\,dx = \left[\frac{x}{2}+\frac{\sin 2x}{4}\right]_0^{\pi} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

5. $\int_0^{\pi/4}\cos 2x\,dx = \left[\frac{\sin 2x}{2}\right]_0^{\pi/4} = \frac{\sin(\pi/2)}{2} = \frac{1}{2}$.

Q1 (2 marks): $\sin^2 4x = \frac{1-\cos 8x}{2}$ [1]. $\int \sin^2 4x\,dx = \frac{x}{2} - \frac{\sin 8x}{16} + C$ [1].

Q2 (3 marks): $\cos^2 3x = \frac{1+\cos 6x}{2}$ [1]. Antiderivative: $\frac{x}{2}+\frac{\sin 6x}{12}$ [1]. At $x=\pi/6$: $\frac{\pi}{12}+\frac{\sin\pi}{12} = \frac{\pi}{12}+0 = \frac{\pi}{12}$. At $x=0$: $0$. Answer: $\frac{\pi}{12}$ [1].

Q3 (3 marks): On $[0,\pi/4]$, $\cos^2 x \geq \sin^2 x$; on $[\pi/4,\pi/2]$, $\sin^2 x \geq \cos^2 x$ [1]. By symmetry, each sub-area equals $\frac{1}{2}$ (from worked example). Total area $= 2 \times \frac{1}{2} = 1$ [2].

Boss battle · The Trig Identity Titan

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Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering $\sin^2$ and $\cos^2$ integration questions. Lighter alternative to the boss.

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← Lesson 5 · Algebraic Manipulation Before Integration Lesson 7 · Integrating Powers of Sine and Cosine →

Module overview · Extension 1 · Checkpoint 2