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11.5 Integration by Substitution

11.5 Integration by Substitution

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Question 36
a.

Use the substitution x=3sin⁡ux = 3 \sin ux=3sinu to show that

∫01.52x+3(9−x2)32 dx=∫0p(23sec⁡utan⁡u+13sec⁡2u) du \int_{0}^{1.5} \frac{2x+3}{(9-x^2)^{\frac{3}{2}}} \, dx = \int_{0}^{p} \left( \frac{2}{3} \sec u \tan u + \frac{1}{3} \sec^2 u \right) \, du ∫01.5​(9−x2)23​2x+3​dx=∫0p​(32​secutanu+31​sec2u)du

where p p\,p is a constant to be found.

[5]
b.

Hence find the exact value of

∫01.52x+3(9−x2)32 dx \int_{0}^{1.5} \frac{2x+3}{(9-x^2)^{\frac{3}{2}}} \, dx ∫01.5​(9−x2)23​2x+3​dx
[3]
Markscheme

11.5 Integration by Substitution Questions

  1. A Level
  2. /Maths
  3. /11.5 Integration by Substitution

59 exam-style questions on Edexcel A Level Maths 11.5 Integration by Substitution. Each one has a worked solution and a mark scheme showing where the marks go.

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