A researcher models the radial gradient of a potential field V V\,V using the equation dVdr=1r2r2−16\displaystyle \frac{dV}{dr} = \frac{1}{r^2 \sqrt{r^2 - 16}}drdV=r2r2−161 for r>4r > 4r>4. To solve for VVV, the researcher considers the transformation y=secϕy = \sec \phiy=secϕ.
(i) Express y y\,y in terms of cosϕ\cos \phicosϕ.
(ii) Hence, show that dydϕ=secϕtanϕ\displaystyle \frac{dy}{d\phi} = \sec \phi \tan \phidϕdy=secϕtanϕ.
(iii) Show that for 0<ϕ<π2\displaystyle 0 < \phi < \frac{\pi}{2}0<ϕ<2π, y2−1y=sinϕ\displaystyle \frac{\sqrt{y^2-1}}{y} = \sin \phiyy2−1=sinϕ.
(i) Use the substitution r=4secϕr = 4 \sec \phir=4secϕ to show that for r>4r > 4r>4, the integral ∫1r2r2−16 dr\displaystyle \int \frac{1}{r^2 \sqrt{r^2 - 16}} \, dr∫r2r2−161dr can be written as k∫cosϕ dϕ k \int \cos \phi \, d\phi\,k∫cosϕdϕ where k k\,k is a constant to be determined.
(ii) Hence, show that ∫1r2r2−16 dr=r2−1616r+C\displaystyle \int \frac{1}{r^2 \sqrt{r^2 - 16}} \, dr = \frac{\sqrt{r^2 - 16}}{16r} + C∫r2r2−161dr=16rr2−16+C.
864 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.