IB Questionbank

Updates to Questionbank

User interface language: English | Español

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

[N/A]

Syllabus sections

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

Show 27 related questions

19M.2.AHL.TZ1.H_5:
Use integration by parts to find $\int {{{\left( {{\text{ln}}\,x} \right)}^2}} {\text{d}}x$ .
EXM.3.AHL.TZ0.3c:
Hence, find the exact values of ${I_3}$ and ${I_4}$ .
EXM.3.AHL.TZ0.3b.ii:
Explain where the condition $n \geqslant 2$ was used in your proof.
18N.1.AHL.TZ0.H_10b:
Hence, show that $\int {{{\text{e}}^x}\,{\text{cos}}{\,^2}x{\text{d}}x = } \frac{{{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{{10}}{\text{cos}}\,2x + \frac{{{{\text{e}}^x}}}{2} + c$ , $c \in \mathbb{R}$ .
EXM.3.AHL.TZ0.3d:
Use the substitution $x = \frac{\pi }{2} - u$ to show that ${J_n} = {I_n}$ .
EXM.3.AHL.TZ0.3e:
Hence, find the exact values of ${J_{5}}$ and ${J_{6}}$
18N.1.AHL.TZ0.H_10a:
Use integration by parts to show that $\int {{{\text{e}}^x}\,{\text{cos}}\,2x{\text{d}}x = } \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x + c$ , $c \in \mathbb{R}$ .
19M.1.AHL.TZ2.H_4:
Using the substitution $u = {\text{sin}}\,x$ , find $\int {\frac{{{\text{co}}{{\text{s}}^3}x\,{\text{d}}x}}{{\sqrt {{\text{sin}}\,x} }}}$ .
22M.1.AHL.TZ1.7b:
Find $E (X)$ .
22M.1.AHL.TZ2.7:
By using the substitution $u = sec x$ or otherwise, find an expression for $\int_{0}^{\frac{π}{3}} {sec}^{n} x \tan x d x$ in terms of $n$ , where $n$ is a non-zero real number.
17M.1.AHL.TZ2.H_6b:
Hence find the value of $\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x}$ .
EXM.3.AHL.TZ0.3f:
Find the exact values of ${T_{0}}$ and ${T_{1}}$ .
18M.1.AHL.TZ2.H_8b:
Hence find the value of $\frac{1}{2}\int\limits_1^9 {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}}$ , expressing your answer in the form arctan $q$ , where $q \in \mathbb{Q}$ .
EXM.3.AHL.TZ0.3g.ii:
Explain where the condition $n \geqslant 2$ was used in your proof.
17M.1.AHL.TZ2.H_6a:
Using the substitution $x = \tan \theta$ show that $\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x = } \int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta }$ .
18N.1.AHL.TZ0.H_10d:
Find the area enclosed by the curve and the $x$ -axis between B and D, as shaded on the diagram.
EXM.3.AHL.TZ0.3h:
Hence, find the exact values of ${T_{2}}$ and ${T_{3}}$ .
EXM.3.AHL.TZ0.3g.i:
Use the fact that ${\text{ta}}{{\text{n}}^2}x = {\text{se}}{{\text{c}}^2}x - 1$ to show that ${T_n} = \frac{1}{{n - 1}} - {T_{n - 2}}{\text{,}}\,\,n \geqslant 2$ .
EXM.3.AHL.TZ0.3b.i:
Use integration by parts to show that ${I_n} = \frac{{n - 1}}{n}{I_{n - 2}}{\text{,}}\,\,n \geqslant 2$ .
EXM.3.AHL.TZ0.3a:
Find the exact values of ${I_0}$ , ${I_1}$ and ${I_2}$ .
22M.1.AHL.TZ1.7a:
Find the value of $k$ .
21N.1.AHL.TZ0.8:
Solve the differential equation $\frac{d y}{d x} = \frac{\ln 2 x}{x^{2}} - \frac{2 y}{x}, x > 0$ , given that $y = 4$ at $x = \frac{1}{2}$ .

Give your answer in the form $y = f (x)$ .
18M.1.AHL.TZ2.H_8a:
Use the substitution $u = {x^{\frac{1}{2}}}$ to find $\int {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}}$ .
18N.1.AHL.TZ0.H_10c:
Find the $x$ -coordinates of A and of C , giving your answers in the form $a + {\text{arctan}}\,b$ , where $a$ , $b \in \mathbb{R}$ .
17M.1.AHL.TZ1.H_9:
Find $\int {\arcsin x\,{\text{d}}x}$
21M.1.AHL.TZ2.9:
By using the substitution $u = \sin x$ , find $\int \frac{\sin x \cos x}{\sin^{2} x - \sin x - 2} d x$ .
18M.1.AHL.TZ2.H_6b:
Hence, or otherwise, find $\int\limits_0^\pi {{{\text{e}}^{ - x}}\,{\text{sin}}\,x\,{\text{d}}x}$ .

Hide related questions

Topic 5 —Calculus