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Date	May 2017	Marks available	3	Reference code	17M.1.AHL.TZ2.H_6
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Find and Hence	Question number	H_6	Adapted from	N/A

Question

Using the substitution $x = \tan \theta$ show that $\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x = } \int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta }$ .

[4]

Hence find the value of $\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x}$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

let $x = \tan \theta$

$\Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}\theta }} = {\sec ^2}\theta$ (A1)

$\int {\frac{1}{{{{({x^2} + 1)}^2}}}{\text{d}}x = \int {\frac{{{{\sec }^2}\theta }}{{{{({{\tan }^2}\theta + 1)}^2}}}{\text{d}}\theta } }$ M1

Note: The method mark is for an attempt to substitute for both $x$ and ${\text{d}}x$ .

$= \int {\frac{1}{{{{\sec }^2}\theta }}{\text{d}}\theta }$ (or equivalent) A1

when $x = 0,{\text{ }}\theta = 0$ and when $x = 1,{\text{ }}\theta = \frac{\pi }{4}$ M1

$\int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta }$ AG

[4 marks]

$\left( {\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x} = \int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta } } \right) = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\left( {1 + \cos 2\theta } \right){\text{d}}\theta }$ M1

$= \frac{1}{2}\left[ {\theta + \frac{{\sin 2\theta }}{2}} \right]_0^{\frac{\pi }{4}}$ A1

$= \frac{\pi }{8} + \frac{1}{4}$ A1

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » AHL 5.16—Integration by substitution, parts and repeated parts

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EXM.3.AHL.TZ0.3h:
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