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 ds W = \int_{0}^{2} \frac{3s+4}{(16-s^2)^{\frac{3}{2}}} \, ds W=∫02(16−s2)233s+4dsUse the substitution s=4sinθs = 4 \sin \thetas=4sinθ to show that
∫023s+4(16−s2)32 ds=∫0p(34secθtanθ+14sec2θ) 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)233s+4ds=∫0p(43secθtanθ+41sec2θ)dθwhere ppp is a constant to be found.
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)233s+4ds119 exam-style questions on AQA A Level Maths 1.11.5 Integration by substitution and by parts (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.