2016 H2 Maths Paper 2 Q2 — Worked Solution

Part (a)(i)

3 marks

Approach

$x^{2}\cos nx$ is a product of a polynomial and a trigonometric function, so we integrate by parts, choosing $x^{2}$ as the part that differentiates to something simpler and applying the process twice.

Let $u = x^{2}$ and $\frac{dv}{dx} = \cos nx$ , so $\frac{du}{dx} = 2x$ and $v = \frac{1}{n}\sin nx$ .

\int x^{2}\cos nx\, dx

= \frac{x^{2}}{n}\sin nx - \int\frac{2x}{n}\sin nx\, dx

For $\int x\sin nx\, dx$ , integrate by parts again with $u = x$ , $\frac{dv}{dx} = \sin nx$ , so $\frac{du}{dx} = 1$ , $v = - \frac{1}{n}\cos nx$ :

\int x\sin nx\, dx

= - \frac{x}{n}\cos nx + \int\frac{1}{n}\cos nx\, dx

= - \frac{x}{n}\cos nx + \frac{1}{n^{2}}\sin nx

Substituting back:

\int x^{2}\cos nx\, dx

= \frac{x^{2}}{n}\sin nx - \frac{2}{n}\left( - \frac{x}{n}\cos nx + \frac{1}{n^{2}}\sin nx \right) + C

= \frac{x^{2}}{n}\sin nx + \frac{2x}{n^{2}}\cos nx - \frac{2}{n^{3}}\sin nx + C

$\therefore\int x^{2}\cos nx\, dx = \frac{x^{2}}{n}\sin nx + \frac{2x}{n^{2}}\cos nx - \frac{2}{n^{3}}\sin nx + C$ .

Part (a)(ii)

2 marks

Approach

We evaluate the antiderivative from (i) at $x = 2\pi$ and $x = \pi$ . Since $n$ is an integer, $\sin n\pi = \sin 2n\pi = 0$ for every $n$ , so only the $\cos nx$ term survives; but $\cos 2n\pi = 1$ always, while $\cos n\pi = ( - 1)^{n}$ depends on the parity of $n$ , so the two cases must be treated separately.

Using the result of (i):

\int_{\pi}^{2\pi}x^{2}\cos nx\, dx = \left\lbrack \frac{x^{2}}{n}\sin nx + \frac{2x}{n^{2}}\cos nx - \frac{2}{n^{3}}\sin nx \right\rbrack_{\pi}^{2\pi}

Since $n$ is an integer, $\sin 2n\pi = \sin n\pi = 0$ , so this reduces to

\int_{\pi}^{2\pi}x^{2}\cos nx\, dx = \frac{4\pi}{n^{2}}\cos 2n\pi - \frac{2\pi}{n^{2}}\cos n\pi

= \frac{4\pi}{n^{2}} - \frac{2\pi}{n^{2}}\cos n\pi

Note: $\cos n\pi = ( - 1)^{n}$ , which is $1$ when $n$ is even and $- 1$ when $n$ is odd.

If $n$ is even:

\int_{\pi}^{2\pi}x^{2}\cos nx\, dx = \frac{4\pi}{n^{2}} - \frac{2\pi}{n^{2}} = \frac{2\pi}{n^{2}}

If $n$ is odd:

\int_{\pi}^{2\pi}x^{2}\cos nx\, dx = \frac{4\pi}{n^{2}} + \frac{2\pi}{n^{2}} = \frac{6\pi}{n^{2}}

$\therefore\int_{\pi}^{2\pi}x^{2}\cos nx\, dx = a\frac{\pi}{n^{2}}$ , where $a = 2$ if $n$ is even, and $a = 6$ if $n$ is odd.

Part (b)

5 marks

Approach

We need $\int_{0}^{2}\pi y^{2}\, dx = \pi\int_{0}^{2}\frac{x^{3}}{\left( 9 - x^{2} \right)^{2}}\, dx$ . The given substitution $u = 9 - x^{2}$ turns $9 - x^{2}$ into $u$ and, since $du = - 2x\, dx$ , turns the leftover $x^{2} \cdot x\, dx$ into $(9 - u)$ times a constant times $du$ — this is why the substitution is chosen: it clears the awkward $x^{2}$ inside the bracket as well as supplying the $x\, dx$ .

Let $y = \frac{x\sqrt{x}}{9 - x^{2}}$ , so $y^{2} = \frac{x^{3}}{\left( 9 - x^{2} \right)^{2}}$ .

Volume $\text{=}\pi\int_{0}^{2}y^{2}\, dx = \pi\int_{0}^{2}\frac{x^{3}}{\left( 9 - x^{2} \right)^{2}}\, dx$ .

Let $u = 9 - x^{2}$ . Then $\frac{du}{dx} = - 2x$ , so $x\, dx = - \frac{1}{2}\, du$ , and $x^{2} = 9 - u$ .

When $x = 0$ , $u = 9$ ; when $x = 2$ , $u = 5$ .

\int_{0}^{2}\frac{x^{3}}{\left( 9 - x^{2} \right)^{2}}\, dx = \int_{0}^{2}\frac{x^{2}}{\left( 9 - x^{2} \right)^{2}} \cdot x\, dx

= \int_{9}^{5}\frac{9 - u}{u^{2}}\left( - \frac{1}{2} \right)du

= \frac{1}{2}\int_{5}^{9}\left( \frac{9}{u^{2}} - \frac{1}{u} \right)du

= \frac{1}{2}\left\lbrack - \frac{9}{u} - \ln u \right\rbrack_{5}^{9}

Evaluating:

\frac{1}{2}\left\lbrack - \frac{9}{u} - \ln u \right\rbrack_{5}^{9}

= \frac{1}{2}\left\lbrack \left( - 1 - \ln 9 \right) - \left( - \frac{9}{5} - \ln 5 \right) \right\rbrack

= \frac{1}{2}\left\lbrack \frac{4}{5} + \ln 5 - \ln 9 \right\rbrack

= \frac{1}{2}\left( \frac{4}{5} + \ln\frac{5}{9} \right)

= \frac{2}{5} - \frac{1}{2}\ln\frac{9}{5}

So Volume $\text{=}\pi\left( \frac{2}{5} - \frac{1}{2}\ln\frac{9}{5} \right)$ .

To check your answer: numerically this is

\pi(0.4 - 0.2939\ldots) = \pi(0.1061\ldots) \approx 0.333

, which is positive and small, consistent with a thin sliver of solid swept from a curve with

y(0) = 0

and

y(2) = \frac{2\sqrt{2}}{5} \approx 0.566

. ✓

$\therefore$ the exact volume is $\pi\left( \frac{2}{5} - \frac{1}{2}\ln\frac{9}{5} \right)\ \text{units}^{3}$ .

2016 H2 Maths Paper 2, Question 2

Part (a)(i)

Part (a)(ii)

Part (b)

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