A researcher is studying the intensity of light propagation through a specific lens assembly. The calculation of the phase shift involves the integral:
I=∫1r2r2−16 drI = \int \frac{1}{r^2 \sqrt{r^2 - 16}} \, drI=∫r2r2−161dr
Consider the variable transformation v=secϕv = \sec \phiv=secϕ.
(i) Express vvv in terms of cosϕ\cos \phicosϕ.
(ii) Hence, show that dvdϕ=secϕtanϕ\frac{dv}{d\phi} = \sec \phi \tan \phidϕdv=secϕtanϕ.
(iii) Prove that for 0<ϕ<π20 < \phi < \frac{\pi}{2}0<ϕ<2π, v2−1v=sinϕ\frac{\sqrt{v^2-1}}{v} = \sin \phivv2−1=sinϕ.
(i) Use the substitution r=4secϕr = 4 \sec \phir=4secϕ to show that for r>4r > 4r>4, the integral III can be expressed as:
I=k∫cosϕ dϕI = k \int \cos \phi \, d\phiI=k∫cosϕdϕ
where kkk is a constant to be found.
(ii) Hence, show that
∫1r2r2−16 dr=r2−1616r+C\int \frac{1}{r^2 \sqrt{r^2 - 16}} \, dr = \frac{\sqrt{r^2 - 16}}{16r} + C∫r2r2−161dr=16rr2−16+C
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.