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+C864 exam-style questions on Edexcel A Level Maths Integration, covering 11.1 Integrating Standard Functions, 11.2 Integrating f(ax + b), 11.3 Using Trigonometric Identities, 11.4 Reverse Chain Rule, 11.5 Integration by Substitution, 11.6 Integration by Parts, 11.7 Partial Fractions, 11.8 Finding Areas, 11.9 The Trapezium Rule, 11.10 Solving Differential Equations, and 11.11 Modelling with Differential Equations. Each one has a worked solution and a mark scheme showing where the marks go.