Year 12 Maths Extension 1 (2027) Calculus

Integration of sin²nx and cos²nx

Q: How do you integrate sin squared x?

Rewrite sin squared x as one half of one minus cos two x, then integrate to get x over two minus one quarter sin two x plus C.

Q: How do you integrate cos squared x?

Rewrite cos squared x as one half of one plus cos two x, giving x over two plus one quarter sin two x plus C.

Q: What identity do you use?

The double-angle identities: sin squared n x equals one half of one minus cos two n x, and cos squared n x equals one half of one plus cos two n x.

Q: Why can't you integrate sin squared directly?

There is no elementary antiderivative in that form; the identity turns it into a constant plus a cosine, which do integrate.

Q: What changes for sin squared of n x?

The identity uses cos two n x, and integrating that term gives one over two n times sin two n x.

20 practice questions 2 video lessons Theory + worked examples

Learn to integrate squared trigonometric functions for NSW Year 12 Mathematics Extension 1. Expressions such as sin squared and cos squared cannot be integrated directly, so a double-angle identity is used to rewrite them first.

You will learn to prove and apply the power-reduction identities, then integrate the resulting expressions — a standard technique that appears in area and volume problems throughout the Extension 1 course.

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Practice questions

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Watch 2 video(s)

Using trig identities for integration Watch
The Integration of sin^2 x and cos^2 x Watch

Theory

Squared trig functions cannot be integrated directly — rewrite with a double-angle identity first: $\sin^{2} n x = \frac{1}{2} (1 - \cos 2 n x)$ , $\cos^{2} n x = \frac{1}{2} (1 + \cos 2 n x)$ . This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-04) then integrates term by term.

Squared trig functions like $\sin^{2} n x$ and $\cos^{2} n x$ cannot be integrated directly — first rewrite them with a double-angle identity, then integrate term by term.

$\sin^{2} n x = \frac{1}{2} (1 - \cos 2 n x)$ and $\cos^{2} n x = \frac{1}{2} (1 + \cos 2 n x)$ . These come from $\cos 2 n x = 1 - 2 \sin^{2} n x = 2 \cos^{2} n x - 1$ .

Once rewritten, each piece is a constant or a cosine, both easy to integrate.

NESA link. Part of the Year 12 Calculus topic, outcome ME1-12-04 ("selects and applies differentiation and integration techniques to solve problems") with MAO-WM-01. This is part of the Techniques of integration focus area.

\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)

oscillates about its average

\frac{1}{2}

\int_{0}^{π / 2} \cos^{2} x d x = \frac{π}{4}

\sin^{2} n x = \frac{1}{2} (1 - \cos 2 n x), \cos^{2} n x = \frac{1}{2} (1 + \cos 2 n x) .

sin^2(nx) = (1/2)(1 - cos 2nx); cos^2(nx) = (1/2)(1 + cos 2nx)

Integrating term by term:

\int \sin^{2} n x d x = \frac{x}{2} - \frac{1}{4 n} \sin 2 n x + C, \int \cos^{2} n x d x = \frac{x}{2} + \frac{1}{4 n} \sin 2 n x + C .

integral sin^2(nx) dx = x/2 - (1/(4n)) sin 2nx + C; integral cos^2(nx) dx = x/2 + (1/(4n)) sin 2nx + C

Watch the angle. The identity for $\sin^{2} n x$ uses $\cos 2 n x$ (double the inside angle), and integrating $\cos 2 n x$ brings a $\frac{1}{2 n}$ .

How to integrate $\sin^{2} n x$ or $\cos^{2} n x$

Apply the identity: $\sin^{2} n x = \frac{1}{2} (1 - \cos 2 n x)$ or $\cos^{2} n x = \frac{1}{2} (1 + \cos 2 n x)$ .
Split the integral into the constant term and the cosine term.
Integrate, remembering $\int \cos 2 n x d x = \frac{1}{2 n} \sin 2 n x$ .
For a definite integral, substitute the limits (the $\sin$ term often vanishes at nice angles).

Example 1 — Indefinite (cosine)

Find

\int \cos^{2} 4 x d x

Solution

$\cos^{2} 4 x = \frac{1}{2} (1 + \cos 8 x)$ .

$\int \cos^{2} 4 x d x$	$=$	$\frac{1}{2} (x + \frac{1}{8} \sin 8 x) + C$
	$=$	$\frac{x}{2} + \frac{1}{16} \sin 8 x + C$

integral cos^2(4x) dx = x/2 + (1/16) sin 8x + C

So the integral is $\frac{x}{2} + \frac{1}{16} \sin 8 x + C$ .

Example 2 — Definite (cosine)

Evaluate

\int_{0}^{π / 2} \cos^{2} x d x

Solution

$\cos^{2} x = \frac{1}{2} (1 + \cos 2 x)$ .

$\int_{0}^{π / 2} \cos^{2} x d x$	$=$	$\frac{1}{2} [x + \frac{1}{2} \sin 2 x]_{0}^{π / 2}$
	$=$	$\frac{1}{2} \cdot \frac{π}{2} = \frac{π}{4}$

integral 0 to pi/2 cos^2 x dx = pi/4

Value: $\frac{π}{4}$ .

Example 3 — Definite (sine)

Evaluate

\int_{0}^{π / 6} \sin^{2} x d x

Solution

$\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)$ .

$\int_{0}^{π / 6} \sin^{2} x d x$	$=$	$\frac{1}{2} [x - \frac{1}{2} \sin 2 x]_{0}^{π / 6}$
	$=$	$\frac{π}{12} - \frac{\sqrt{3}}{8}$

integral 0 to pi/6 sin^2 x dx = pi/12 - sqrt(3)/8

Value: $\frac{π}{12} - \frac{\sqrt{3}}{8}$ .

Example 4 — A multiple angle

Evaluate

\int_{0}^{π / 4} \sin^{2} 2 x d x

Solution

$\sin^{2} 2 x = \frac{1}{2} (1 - \cos 4 x)$ .

$\int_{0}^{π / 4} \sin^{2} 2 x d x$	$=$	$\frac{1}{2} [x - \frac{1}{4} \sin 4 x]_{0}^{π / 4}$
	$=$	$\frac{1}{2} \cdot \frac{π}{4} = \frac{π}{8}$

integral 0 to pi/4 sin^2(2x) dx = pi/8

Value: $\frac{π}{8}$ .

Common pitfalls

Integrating directly.

\sin^{2}

and

\cos^{2}

must be rewritten with a double-angle identity first — you cannot integrate them as they stand.

The doubled angle. For

\sin^{2} n x

the identity uses

\cos 2 n x

; integrating it brings a

\frac{1}{2 n}

Sign of the identity.

\sin^{2}

uses

1 - \cos 2 n x

;

\cos^{2}

uses

1 + \cos 2 n x

Dropping the half. Keep the leading

\frac{1}{2}

from the identity throughout.

Frequently asked questions

How do you integrate sin squared x?

Rewrite $\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)$ , then integrate to get $\frac{x}{2} - \frac{1}{4} \sin 2 x + C$ .

How do you integrate cos squared x?

Rewrite $\cos^{2} x = \frac{1}{2} (1 + \cos 2 x)$ , giving $\frac{x}{2} + \frac{1}{4} \sin 2 x + C$ .

What identity do you use?

The double-angle identities $\sin^{2} n x = \frac{1}{2} (1 - \cos 2 n x)$ and $\cos^{2} n x = \frac{1}{2} (1 + \cos 2 n x)$ .

Why can't you integrate sin squared directly?

There is no elementary antiderivative of $\sin^{2}$ in that form; the identity turns it into a constant plus a cosine, which do integrate.

What changes for sin squared of nx?

The identity uses $\cos 2 n x$ , and integrating that term gives $\frac{1}{2 n} \sin 2 n x$ .

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Integration of sin²nx and cos²nx

📖 Theory

How to integrate sin2⁡nx or cos2⁡nx

Common pitfalls

Frequently asked questions

Theory

How to integrate $\sin^{2} n x$ or $\cos^{2} n x$