2023 H2 Maths Paper 1 Q4 — Worked Solution

Part (a)

2 marks

Using the product-to-sum identity $\cos A\cos B = \frac{1}{2}\left\lbrack \cos(A - B) + \cos(A + B) \right\rbrack$ with $A = px$ , $B = qx$ :

\cos(px)\cos(qx) = \frac{1}{2}\left\lbrack \cos\left( (p - q)x \right) + \cos\left( (p + q)x \right) \right\rbrack

Since $p \neq q$ and $p \neq - q$ , both $p - q$ and $p + q$ are non-zero, so integrating term by term gives sine terms throughout:

\int\cos(px)\cos(qx)\, dx

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

= \frac{1}{2}\left\lbrack \frac{\sin\left( (p - q)x \right)}{p - q} + \frac{\sin\left( (p + q)x \right)}{p + q} \right\rbrack + c

To check your answer: differentiating,

\frac{1}{2}\left\lbrack \cos\left( (p - q)x \right) + \cos\left( (p + q)x \right) \right\rbrack = \cos(px)\cos(qx)

by the identity used ✓

$\therefore$ $\int\cos(px)\cos(qx)\, dx = \frac{\sin\left( (p - q)x \right)}{2(p - q)} + \frac{\sin\left( (p + q)x \right)}{2(p + q)} + c$ .

Part (b)

8 marks

Integrating by parts with $u = x$ and $dv = \cos(nx)\, dx$ , so $du = dx$ and $v = \frac{\sin(nx)}{n}$ :

\int x\cos(nx)\, dx = x \cdot \frac{\sin(nx)}{n} - \int\frac{\sin(nx)}{n}\, dx

= \frac{x\sin(nx)}{n} - \frac{1}{n}\left( - \frac{\cos(nx)}{n} \right) + c

= \frac{x\sin(nx)}{n} + \frac{\cos(nx)}{n^{2}} + c

$\therefore$ $\int x\cos(nx)\, dx = \frac{x\sin(nx)}{n} + \frac{\cos(nx)}{n^{2}} + c$ (shown).

Part (c)

Using the result of (b), we evaluate at the limits $x = 0$ and $x = \pi$ :

\int_{0}^{\pi}x\cos(nx)\, dx = \left\lbrack \frac{x\sin(nx)}{n} + \frac{\cos(nx)}{n^{2}} \right\rbrack_{0}^{\pi}

Since $n$ is a positive integer, $\sin(n\pi) = 0$ and $\cos(n\pi) = ( - 1)^{n}$ , while at $x = 0$ the bracket equals $\frac{1}{n^{2}}$ :

= \left( 0 + \frac{( - 1)^{n}}{n^{2}} \right) - \left( 0 + \frac{1}{n^{2}} \right)

= \frac{( - 1)^{n} - 1}{n^{2}}

To check your answer: for

n = 1

,

\int_{0}^{\pi}x\cos x\, dx = \left\lbrack x\sin x + \cos x \right\rbrack_{0}^{\pi} = (0 - 1) - (0 + 1) = - 2

, matching

\frac{( - 1)^{1} - 1}{1^{2}} = - 2

✓

If $n$ is even, $( - 1)^{n} = 1$ , so the integral equals $0$ . If $n$ is odd, $( - 1)^{n} = - 1$ , so the integral equals $- \frac{2}{n^{2}}$ .

$\therefore$ $\int_{0}^{\pi}x\cos(nx)\, dx = \frac{k}{n^{2}}$ , with $k = 0$ for even $n$ and $k = - 2$ for odd $n$ .

Part (d)

Approach

We first need to know where $\cos(2x)$ changes sign on $\left\lbrack 0,\frac{\pi}{2} \right\rbrack$ in order to remove the modulus, then apply the antiderivative from (b) with $n = 2$ separately on each piece.

On $\left\lbrack 0,\frac{\pi}{2} \right\rbrack$ , $\cos(2x) = 0$ at $x = \frac{\pi}{4}$ , with $\cos(2x) > 0$ for $x < \frac{\pi}{4}$ and $\cos(2x) < 0$ for $x > \frac{\pi}{4}$ . So $\left| x\cos(2x) \right| = x\cos(2x)$ on $\left\lbrack 0,\frac{\pi}{4} \right\rbrack$ and $\left| x\cos(2x) \right| = - x\cos(2x)$ on $\left\lbrack \frac{\pi}{4},\frac{\pi}{2} \right\rbrack$ , giving

\int_{0}^{\pi/2}\left| x\cos(2x) \right|\, dx = \int_{0}^{\pi/4}x\cos(2x)\, dx - \int_{\pi/4}^{\pi/2}x\cos(2x)\, dx

Using the result of (b) with $n = 2$ , let $F(x) = \frac{x\sin(2x)}{2} + \frac{\cos(2x)}{4}$ . Then

F(0) = \frac{1}{4},\quad F\left( \frac{\pi}{4} \right) = \frac{\pi}{8},\quad F\left( \frac{\pi}{2} \right) = - \frac{1}{4}

so

\int_{0}^{\pi/4}x\cos(2x)\, dx = F\left( \frac{\pi}{4} \right) - F(0) = \frac{\pi}{8} - \frac{1}{4}

\int_{\pi/4}^{\pi/2}x\cos(2x)\, dx = F\left( \frac{\pi}{2} \right) - F\left( \frac{\pi}{4} \right) = - \frac{1}{4} - \frac{\pi}{8}

Subtracting:

\int_{0}^{\pi/2}\left| x\cos(2x) \right|\, dx = \left( \frac{\pi}{8} - \frac{1}{4} \right) - \left( - \frac{1}{4} - \frac{\pi}{8} \right)

= \frac{\pi}{4}

To check your answer:

\frac{\pi}{4} \approx 0.785

, matching a numerical estimate of

\int_{0}^{\pi/2}\left| x\cos(2x) \right|\, dx

✓

$\therefore$ $\int_{0}^{\pi/2}\left| x\cos(2x) \right|\, dx = \frac{\pi}{4}$ .

2023 H2 Maths Paper 1, Question 4

Part (a)

Part (b)

Part (c)

Part (d)

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