AS & A Level · Practice questions · Integration using trigonometric identities

One identity away.

Six original Cambridge-style questions where an identity turns an awkward trigonometric integrand into standard integrals, with exact answers.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Choose the identity, then integrate term by term.

Squares of sin or cos: use \(\cos 2\theta\) to lower the power; the angle doubles.
\(\tan^2\): replace by \(\sec^2 - 1\).
Products: \(\sin x\cos x = \dfrac{1}{2}\sin 2x\).
Brackets: expand, then use \(\sin^2 + \cos^2 = 1\).
Evaluate in radians and keep \(\pi\) and surds exact.

Recall

[2 marks]

Find \(\displaystyle\int \cos^2 x\,dx\).

\(\cos^2 x = \dfrac{1}{2}(1 + \cos 2x)\). ✓
\(\displaystyle\int \cos^2 x\,dx = \dfrac{x}{2} + \dfrac{1}{4}\sin 2x + c\). ✓

Calculation

[3 marks]

Find the exact value of \(\displaystyle\int_0^{\pi/4} \sin^2 x\,dx\).

\(\displaystyle\int \sin^2 x\,dx = \dfrac{x}{2} - \dfrac{1}{4}\sin 2x\). ✓
Upper: \(\dfrac{\pi}{8} - \dfrac{1}{4}\sin\dfrac{\pi}{2} = \dfrac{\pi}{8} - \dfrac{1}{4}\). Lower: 0. ✓
Answer: \(\dfrac{\pi}{8} - \dfrac{1}{4}\). ✓

Calculation

[3 marks]

Find the exact value of \(\displaystyle\int_0^{\pi/4} \left(2 + \tan^2 x\right) dx\).

\(2 + \tan^2 x = 2 + \sec^2 x - 1 = 1 + \sec^2 x\). ✓
\(\left[x + \tan x\right]_0^{\pi/4} = \dfrac{\pi}{4} + 1\). ✓

Proof

[5 marks]

(a) Show that \(\cos^4 x - \sin^4 x = \cos 2x\).

(b) Hence find the exact value of \(\displaystyle\int_0^{\pi/8} \left(\cos^4 x - \sin^4 x\right) dx\).

(a) Difference of two squares: \(\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)\). ✓
\(\cos^2 x + \sin^2 x = 1\) and \(\cos^2 x - \sin^2 x = \cos 2x\), so the expression equals \(\cos 2x\). ✓
(b) \(\displaystyle\int_0^{\pi/8} \cos 2x\,dx = \left[\dfrac{1}{2}\sin 2x\right]_0^{\pi/8} = \dfrac{1}{2}\sin\dfrac{\pi}{4}\). ✓
\(= \dfrac{1}{2}\times\dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{2}}{4}\). ✓

Calculation

[4 marks]

Find the exact value of \(\displaystyle\int_0^{\pi/2} \sin^2 x\cos^2 x\,dx\).

\(\sin x\cos x = \dfrac{1}{2}\sin 2x\), so \(\sin^2 x\cos^2 x = \dfrac{1}{4}\sin^2 2x\). ✓
\(\sin^2 2x = \dfrac{1}{2}(1 - \cos 4x)\), so the integrand is \(\dfrac{1}{8}(1 - \cos 4x)\). ✓
\(\left[\dfrac{x}{8} - \dfrac{1}{32}\sin 4x\right]_0^{\pi/2} = \dfrac{\pi}{16} - \dfrac{1}{32}\sin 2\pi = \dfrac{\pi}{16}\). ✓

Analysis

[5 marks]

The region under the curve \(y = \sin x\) between \(x = 0\) and \(x = \pi\) is rotated through \(360^\circ\) about the \(x\)-axis to form a solid shaped like a rugby ball.

Find the exact volume of the solid, and give it to 3 significant figures.

Volume \(= \pi\displaystyle\int_0^{\pi} y^2\,dx = \pi\displaystyle\int_0^{\pi} \sin^2 x\,dx\). ✓
\(\sin^2 x = \dfrac{1}{2}(1 - \cos 2x)\), so \(\displaystyle\int_0^{\pi} \sin^2 x\,dx = \left[\dfrac{x}{2} - \dfrac{1}{4}\sin 2x\right]_0^{\pi} = \dfrac{\pi}{2}\). ✓
Volume \(= \pi\times\dfrac{\pi}{2} = \dfrac{\pi^2}{2}\). ✓
\(\dfrac{\pi^2}{2} = 4.93\) cubic units (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.