2017 H2 Maths Paper 1 Q7 — Worked Solution

Part (i)

3 marks

Approach

A product of two sines has no elementary antiderivative on its own; we first rewrite it as a sum using the factor formula $\sin A\sin B = \frac{1}{2}\left\lbrack \cos(A - B) - \cos(A + B) \right\rbrack$ , which is easy to integrate term by term.

Using the factor formula with $A = 2mx$ , $B = 2nx$ :

\sin 2mx\sin 2nx = \frac{1}{2}\left\lbrack \cos\left( 2(m - n)x \right) - \cos\left( 2(m + n)x \right) \right\rbrack

Since $m \neq n$ , both $m - n$ and $m + n$ are non-zero, so we may integrate term by term:

\int\sin 2mx\sin 2nx\, dx

= \frac{1}{2}\int\left\lbrack \cos\left( 2(m - n)x \right) - \cos\left( 2(m + n)x \right) \right\rbrack dx

= \frac{\sin\left( 2(m - n)x \right)}{4(m - n)} - \frac{\sin\left( 2(m + n)x \right)}{4(m + n)} + C

$\therefore\ \int\sin 2mx\sin 2nx\, dx = \frac{\sin\left( 2(m - n)x \right)}{4(m - n)} - \frac{\sin\left( 2(m + n)x \right)}{4(m + n)} + C$

Part (ii)

5 marks

Approach

Expanding $\left( f(x) \right)^{2}$ gives two squared-sine terms and a cross term proportional to $\sin 2mx\sin 2nx$ , which part (i) already handles. Each squared-sine term is dealt with by the double-angle identity $\sin^{2}\theta = \frac{1}{2}\left( 1 - \cos 2\theta \right)$ .

\left( f(x) \right)^{2} = \sin^{2}2mx + 2\sin 2mx\sin 2nx + \sin^{2}2nx

\int_{0}^{\pi}\left( f(x) \right)^{2}\, dx = \int_{0}^{\pi}\sin^{2}2mx\, dx + 2\int_{0}^{\pi}\sin 2mx\sin 2nx\, dx + \int_{0}^{\pi}\sin^{2}2nx\, dx

We first find $\int_{0}^{\pi}\sin^{2}2mx\, dx$ . Using $\sin^{2}2mx = \frac{1}{2}\left( 1 - \cos 4mx \right)$ :

\int_{0}^{\pi}\sin^{2}2mx\, dx

= \frac{1}{2}\left\lbrack x - \frac{\sin 4mx}{4m} \right\rbrack_{0}^{\pi}

= \frac{1}{2}\left( \pi - \frac{\sin 4m\pi}{4m} \right)

Since $m$ is a positive integer, $\sin 4m\pi = 0$ , so

\int_{0}^{\pi}\sin^{2}2mx\, dx = \frac{\pi}{2}

By the same working with $n$ in place of $m$ (and $\sin 4n\pi = 0$ since $n$ is a positive integer),

\int_{0}^{\pi}\sin^{2}2nx\, dx = \frac{\pi}{2}

Now the cross term, using the result of (i):

\int_{0}^{\pi}\sin 2mx\sin 2nx\, dx

= \left\lbrack \frac{\sin\left( 2(m - n)x \right)}{4(m - n)} - \frac{\sin\left( 2(m + n)x \right)}{4(m + n)} \right\rbrack_{0}^{\pi}

Since $m - n$ and $m + n$ are non-zero integers, $\sin\left( 2(m - n)\pi \right) = 0$ and $\sin\left( 2(m + n)\pi \right) = 0$ , and both terms vanish at $x = 0$ too, so

\int_{0}^{\pi}\sin 2mx\sin 2nx\, dx = 0

Combining the three results:

\int_{0}^{\pi}\left( f(x) \right)^{2}\, dx = \frac{\pi}{2} + 2(0) + \frac{\pi}{2}

To check your answer: taking

m = 1

n = 2

as a concrete case and evaluating

\int_{0}^{\pi}\left( \sin 2x + \sin 4x \right)^{2}\, dx

by GC gives

\pi

, which matches. ✓

$\therefore\ \int_{0}^{\pi}\left( f(x) \right)^{2}\, dx = \pi$

2017 H2 Maths Paper 1, Question 7

Part (i)

Part (ii)

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