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Date	November 2017	Marks available	7	Reference code	17N.2.AHL.TZ0.H_8
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Show that	Question number	H_8	Adapted from	N/A

Question

By using the substitution ${x^2} = 2\sec \theta$ , show that $\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} - 4} }} = \frac{1}{4}\arccos \left( {\frac{2}{{{x^2}}}} \right) + c}$ .

Markscheme

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

EITHER

${x^2} = 2\sec \theta$

$2x\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = 2\sec \theta \tan \theta$ M1A1

$\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} - 4} }}}$

$= \int {\frac{{\sec \theta \tan \theta {\text{d}}\theta }}{{2\sec \theta \sqrt {4{{\sec }^2}\theta - 4} }}}$ M1A1

OR

$x = \sqrt 2 {(\sec \theta )^{\frac{1}{2}}}{\text{ }}\left( { = \sqrt 2 {{(\cos \theta )}^{ - \frac{1}{2}}}} \right)$

$\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = \frac{{\sqrt 2 }}{2}{(\sec \theta )^{\frac{1}{2}}}\tan \theta {\text{ }}\left( { = \frac{{\sqrt 2 }}{2}{{(\cos \theta )}^{ - \frac{3}{2}}}\sin \theta } \right)$ M1A1

$\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} - 4} }}}$

$= \int {\frac{{\sqrt 2 {{(\sec \theta )}^{\frac{1}{2}}}\tan \theta {\text{d}}\theta }}{{2\sqrt 2 {{(\sec \theta )}^{\frac{1}{2}}}\sqrt {4{{\sec }^2}\theta - 4} }}{\text{ }}\left( { = \int {\frac{{\sqrt 2 {{(\cos \theta )}^{ - \frac{3}{2}}}\sin \theta {\text{d}}\theta }}{{2\sqrt 2 {{(\cos \theta )}^{ - \frac{1}{2}}}\sqrt {4{{\sec }^2}\theta - 4} }}} } \right)}$ M1A1

THEN

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

$= \frac{1}{4}\int {{\text{d}}\theta }$

$= \frac{\theta }{4} + c$ A1

${x^2} = 2\sec \theta \Rightarrow \cos \theta = \frac{2}{{{x^2}}}$ M1

Note: This M1 may be seen anywhere, including a sketch of an appropriate triangle.

so $\frac{\theta }{4} + c = \frac{1}{4}\arccos \left( {\frac{2}{{{x^2}}}} \right) + c$ AG

[7 marks]

Examiners report

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Syllabus sections

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

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