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 dr I = \int \frac{1}{r^2 \sqrt{r^2 - 16}} \, dr I=∫r2r2−161drConsider 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 dϕ I = k \int \cos \phi \, d\phi I=k∫cosϕdϕwhere 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+C363 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.