2021 H2 Maths Paper 2 Q5 — Worked Solution

Part (a)

2 marks

We use the identity $\tan^{2}\theta = \sec^{2}\theta - 1$ to rewrite the integrand.

\int\tan^{2}5x\ dx

= \int\left( \sec^{2}5x - 1 \right)\ dx

= \frac{1}{5}\tan 5x - x + C

To check your answer: differentiating

\frac{1}{5}\tan 5x - x + C

gives

\sec^{2}5x - 1 = \tan^{2}5x

, the original integrand. ✓

$\therefore$ $\int\tan^{2}5x\ dx = \frac{1}{5}\tan 5x - x + C$ .

Part (b)

3 marks

Approach

the product $\sin 2x\sin 3x$ has no elementary antiderivative as it stands, so we first turn it into a sum of cosines using the product-to-sum identity, which we can integrate term by term.

We use $\sin A\sin B = \frac{1}{2}\left\lbrack \cos(A - B) - \cos(A + B) \right\rbrack$ with $A = 2x$ , $B = 3x$ .

\sin 2x\sin 3x = \frac{1}{2}\left\lbrack \cos(2x - 3x) - \cos(2x + 3x) \right\rbrack

= \frac{1}{2}\left\lbrack \cos( - x) - \cos 5x \right\rbrack

= \frac{1}{2}\left( \cos x - \cos 5x \right)

since $\cos$ is even. Integrating term by term,

\int_{0}^{b}\sin 2x\sin 3x\ dx

= \frac{1}{2}\int_{0}^{b}\left( \cos x - \cos 5x \right)\ dx

= \frac{1}{2}\left\lbrack \sin x - \frac{1}{5}\sin 5x \right\rbrack_{0}^{b}

= \frac{1}{2}\sin b - \frac{1}{10}\sin 5b

(the lower limit contributes $0$ , since $\sin 0 = 0$ ).

To check your answer: at

b = \frac{\pi}{2}

:

\frac{1}{2}\sin\frac{\pi}{2} - \frac{1}{10}\sin\frac{5\pi}{2} = \frac{1}{2}(1) - \frac{1}{10}(1) = 0.4

, matching direct numerical integration of

\sin 2x\sin 3x

from

0

to

\frac{\pi}{2}

. ✓

$\therefore$ $\int_{0}^{b}\sin 2x\sin 3x\ dx = \frac{1}{2}\sin b - \frac{1}{10}\sin 5b$ .

Part (c)

3 marks

Since the derivative of $\ln x$ is $\frac{1}{x}$ , we let $u = \ln x$ , so $du = \frac{1}{x}\ dx$ . When $x = a$ , $u = \ln a$ ; when $x = b$ , $u = \ln b$ .

\int_{a}^{b}\frac{1}{x\ln x}\ dx

= \int_{\ln a}^{\ln b}\frac{1}{u}\ du

= \left\lbrack \ln|u| \right\rbrack_{\ln a}^{\ln b}

= \ln\left( \ln b \right) - \ln\left( \ln a \right)

Since $1 < a < b$ , both $\ln a > 0$ and $\ln b > 0$ , so by the quotient rule for logarithms,

\ln\left( \ln b \right) - \ln\left( \ln a \right) = \ln\left( \frac{\ln b}{\ln a} \right)

To check your answer: with

a = e

,

b = e^{2}

:

\ln\left( \ln b \right) - \ln\left( \ln a \right) = \ln 2 - \ln 1 = \ln 2 \approx 0.693

; direct numerical integration of

\frac{1}{x\ln x}

from

e

to

e^{2}

also gives

\text{≈}0.693

. ✓

$\therefore$ $\int_{a}^{b}\frac{1}{x\ln x}\ dx = \ln\left( \frac{\ln b}{\ln a} \right)$ .

Part (d)

3 marks

We are given $u = 1 + e^{2x}$ , so $\frac{du}{dx} = 2e^{2x}$ , i.e. $e^{2x}\ dx = \frac{1}{2}\ du$ .

\int\frac{e^{2x}}{\left( 1 + e^{2x} \right)^{3}}\ dx

= \int\frac{1}{u^{3}} \cdot \frac{1}{2}\ du

= \frac{1}{2}\int u^{- 3}\ du

= \frac{1}{2}\left( \frac{u^{- 2}}{- 2} \right) + C

= - \frac{1}{4u^{2}} + C

Substituting back $u = 1 + e^{2x}$ ,

- \frac{1}{4u^{2}} + C = - \frac{1}{4\left( 1 + e^{2x} \right)^{2}} + C

To check your answer: differentiating

- \frac{1}{4\left( 1 + e^{2x} \right)^{2}} + C

by the chain rule gives

- \frac{1}{4} \times ( - 2)\left( 1 + e^{2x} \right)^{- 3} \times 2e^{2x} = \frac{e^{2x}}{\left( 1 + e^{2x} \right)^{3}}

, the original integrand. ✓

$\therefore$ $\int\frac{e^{2x}}{\left( 1 + e^{2x} \right)^{3}}\ dx = - \frac{1}{4\left( 1 + e^{2x} \right)^{2}} + C$ .

2021 H2 Maths Paper 2, Question 5

Part (a)

Part (b)

Part (c)

Part (d)

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