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3.6 Integration (A-level only)

3.6 Integration (A-level only)

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Question 287

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

3.6 Integration (A-level only) Questions

  1. A Level
  2. /Maths
  3. /3.6 Integration (A-level only)

363 exam-style questions on CCEA A Level Maths 3.6 Integration (A-level only), covering 3.6.1 Integration (A-level only), 3.6.2 Integration (A-level only), 3.6.3 Integration (A-level only), 3.6.4 Integration (A-level only), 3.6.5 Integration (A-level only), 3.6.6 Integration (A-level only), and 3.6.7 Integration (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.

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