Learn7357Paper 3 · Pure and StatisticsPure content (sections A–J)Integration using trigonometric identities

Pure content (sections A–J) · Subtopic 48 of 66

Integration Using Trigonometric Identities

Some trigonometric expressions can't be integrated as they stand. Rewriting them with identities, such as $\cos^2x=\tfrac12(1+\cos2x)$ and $\tan^2x=\sec^2x-1$ , turns them into standard integrals.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Integrate $\sin^2x$ and $\cos^2x$ using the double-angle formulae.
Integrate $\tan^2x$ using $\tan^2x=\sec^2x-1$ .
Integrate products like $\sin x\cos x$ using $\sin2x=2\sin x\cos x$ .
Evaluate exact definite integrals of these.

Pause and think

Why can't you integrate $\sin^2x$ by "raising the power and dividing", as with $x^2$ ?

Section 2 of 14

Core idea

Start hereThe integral that needs disguising

$\int\sin^2x\,dx$ looks innocent, but there's no standard result for it, and $\frac{\sin^3x}{3}$ is wrong (differentiate it and see). The trick: rewrite $\sin^2x$ using a double-angle formula as something we can integrate.

The big idea

Rewrite before integrating: $\sin^2x=\frac12(1-\cos2x)$ , $\cos^2x=\frac12(1+\cos2x)$ , $\tan^2x=\sec^2x-1$ , $\sin x\cos x=\frac12\sin2x$ .

1. Squares of sine and cosine

\sin^2x

and

\cos^2x

in terms of

\cos2x

$\cos2x=1-2\sin^2x$
Double angle.
$2\sin^2x=1-\cos2x\Rightarrow\sin^2x=\tfrac12(1-\cos2x)$
Rearrange.
$\cos2x=2\cos^2x-1\Rightarrow\cos^2x=\tfrac12(1+\cos2x)$
Likewise.

sin²x is a shifted, squashed cos wave

\sin^2x

wobbles between

0

and

1

around

\tfrac12

, twice as fast as

\sin x

: exactly

\tfrac12-\tfrac12\cos2x

Worked example $\displaystyle\int_0^{\pi/2}\sin^2x\,dx$

Evaluate.

Goal: Exact value.
Strategy: Rewrite, integrate.

$\int_0^{\pi/2}\tfrac12(1-\cos2x)\,dx$
Identity.
$=\left[\tfrac12x-\tfrac14\sin2x\right]_0^{\pi/2}$
$\int\cos2x=\frac12\sin2x$ .
$=\tfrac\pi4-0-0=\tfrac\pi4$
$\sin\pi=0$ .

Answer

\frac\pi4

✓ Check

\sin^2

and

\cos^2

have equal integrals here and sum to

\int1=\frac\pi2

; half is

\frac\pi4

✓.

2. Other identities

Worked example $\int\tan^2x\,dx$ and $\int\sin x\cos x\,dx$

Integrate both.

Goal: Antiderivatives.
Strategy: Identities.

$\int(\sec^2x-1)dx=\tan x-x+c$
$\tan^2=\sec^2-1$ .
$\int\tfrac12\sin2x\,dx=-\tfrac14\cos2x+c$
Double angle.

Answer

\tan x-x+c

;

-\frac14\cos2x+c

✓ CheckDifferentiate

-\frac14\cos2x

\frac12\sin2x=\sin x\cos x

✓.

Worked example $\int(\sin x+\cos x)^2dx$

Integrate.

Goal: Antiderivative.
Strategy: Expand, simplify with identities.

$\sin^2x+2\sin x\cos x+\cos^2x=1+\sin2x$
Pythagoras and double angle.
$\int(1+\sin2x)dx=x-\tfrac12\cos2x+c$
Integrate.

Answer

x-\frac12\cos2x+c

✓ CheckDerivative:

1+\sin2x

✓.

Common misconception $\int\sin^2x\neq\frac{\sin^3x}3$ The power rule only works for powers of

x

(the variable itself), not powers of a function.

Pause and think

$\int_0^\pi\cos^2x\,dx$ ?

Summary

$\sin^2,\cos^2$ via $\cos2x$ .
$\tan^2=\sec^2-1$ .
$\sin x\cos x=\frac12\sin2x$ .

Identity: Rewrites into an integrable form.

HistoryWallis's product. In 1655 John Wallis studied integrals of powers of

\sin x

and found a famous formula:

\dfrac{\pi}{2}=\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{4}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{5}\cdot\dfrac{6}{7}\cdots

. It came from reducing

\int\sin^n x\,dx

step by step with identities.

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