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Question 721

Given that

y=sec⁡θ y = \sec \theta y=secθ
ai.

Express yyy in terms of cos⁡θ\cos \thetacosθ.

[1]
aii.

Hence, show that

dydθ=sec⁡θtan⁡θ\frac{dy}{d\theta} = \sec \theta \tan \theta dθdy​=secθtanθ
[2]
aiii.

Show that for 0<θ<π20 < \theta < \frac{\pi}{2}0<θ<2π​,

y2−1y=sin⁡θ\frac{\sqrt{y^2-1}}{y} = \sin \theta yy2−1​​=sinθ
[2]
bi.

Use the substitution x=3sec⁡ux = 3 \sec ux=3secu to show that for x>3x > 3x>3, the integral

∫1x2x2−9 dx\int \frac{1}{x^2 \sqrt{x^2 - 9}} \, dx ∫x2x2−9​1​dx

can be written as

k∫cos⁡u duk \int \cos u \, du k∫cosudu

where kkk is a constant to be found.

[3]
bii.

Hence, show that

∫1x2x2−9 dx=x2−99x+C\int \frac{1}{x^2 \sqrt{x^2 - 9}} \, dx = \frac{\sqrt{x^2 - 9}}{9x} + C ∫x2x2−9​1​dx=9xx2−9​​+C
[2]
Markscheme

Integration Questions

  1. A Level
  2. /Maths
  3. /Integration

864 exam-style questions on Edexcel A Level Maths Integration, covering 11.1 Integrating Standard Functions, 11.2 Integrating f(ax + b), 11.3 Using Trigonometric Identities, 11.4 Reverse Chain Rule, 11.5 Integration by Substitution, 11.6 Integration by Parts, 11.7 Partial Fractions, 11.8 Finding Areas, 11.9 The Trapezium Rule, 11.10 Solving Differential Equations, and 11.11 Modelling with Differential Equations. Each one has a worked solution and a mark scheme showing where the marks go.

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