11.5 Integration by Substitution
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The work done WWW by a magnetic force on a micro-particle is determined by its displacement sss (in mm). For 0≤s≤20 \le s \le 20≤s≤2, the work required is given by the integral:

W=∫023s+4(16−s2)32 dsW = \int_{0}^{2} \frac{3s+4}{(16-s^2)^{\frac{3}{2}}} \, dsW=∫02​(16−s2)23​3s+4​ds

a.

Use the substitution s=4sin⁡θs = 4 \sin \thetas=4sinθ to show that

∫023s+4(16−s2)32 ds=∫0p(34sec⁡θtan⁡θ+14sec⁡2θ) dθ\int_{0}^{2} \frac{3s+4}{(16-s^2)^{\frac{3}{2}}} \, ds = \int_{0}^{p} \left( \frac{3}{4} \sec \theta \tan \theta + \frac{1}{4} \sec^2 \theta \right) \, d\theta∫02​(16−s2)23​3s+4​ds=∫0p​(43​secθtanθ+41​sec2θ)dθ

where ppp is a constant to be found.

[5]
b.

Hence find the exact value of

∫023s+4(16−s2)32 ds\int_{0}^{2} \frac{3s+4}{(16-s^2)^{\frac{3}{2}}} \, ds∫02​(16−s2)23​3s+4​ds

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11.5 Integration by Substitution Questions

Practise Edexcel A Level Maths 11.5 Integration by Substitution with exam-style questions for A Level Maths. 59 questions, matched to the Edexcel A Level Maths (9MA0) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.

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11.5 Integration by Substitution Questions

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