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Date	May 2021	Marks available	7	Reference code	21M.1.AHL.TZ2.9
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Find	Question number	9	Adapted from	N/A

Question

The lines $l_{1}$ and $l_{2}$ have the following vector equations where $λ, μ \in ℝ$ .

$l_{1} : r_{1} = (\begin{matrix} 3 \\ 2 \\ - 1 \end{matrix}) + λ (\begin{matrix} 2 \\ - 2 \\ 2 \end{matrix})$

$l_{2} : r_{2} = (\begin{matrix} 2 \\ 0 \\ 4 \end{matrix}) + μ (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix})$

By using the substitution $u = \sin x$ , find $\int \frac{\sin x \cos x}{\sin^{2} x - \sin x - 2} d x$ .

Markscheme

$u = \sin x \Rightarrow d u = \cos x d x$ (or equivalent) A1

$= \int \frac{u}{u^{2} - u - 2} d u$ A1

attempt to use partial fractions M1

$(\frac{u}{(u + 1) (u - 2)} \equiv \frac{A}{u + 1} + \frac{B}{u - 2} \Rightarrow u \equiv A (u - 2) + B (u + 1))$

Valid attempt to solve for $A$ and $B$ (M1)

$A = \frac{1}{3}$ and $B = \frac{2}{3}$ A1

$\frac{u}{(u + 1) (u - 2)} \equiv \frac{1}{3 (u + 1)} + \frac{2}{3 (u - 2)}$

$\int (\frac{1}{3 (u + 1)} + \frac{2}{3 (u - 2)}) d u = \frac{1}{3} \ln |u + 1| + \frac{2}{3} \ln |u - 2| (+ C)$ (or equivalent) A1

Note: Condone the absence of $+ C$ or lack of moduli here but not in the final answer.

$= \frac{1}{3} \ln |\sin x + 1| + \frac{2}{3} \ln |\sin x - 2| + C$ A1

Note: Condone further simplification of the correct answer.

[7 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » AHL 5.15—Further derivatives and indefinite integration of these, partial fractions

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