11.5 Integration by Substitution
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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​=r2r2−16​1​ for r>4r > 4r>4. To solve for VVV, the researcher considers the transformation y=sec⁡ϕy = \sec \phiy=secϕ.

a.

(i) Express y y\,y in terms of cos⁡ϕ\cos \phicosϕ.

(ii) Hence, show that dydϕ=sec⁡ϕtan⁡ϕ\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=sin⁡ϕ\displaystyle \frac{\sqrt{y^2-1}}{y} = \sin \phiyy2−1​​=sinϕ.

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b.

(i) Use the substitution r=4sec⁡ϕr = 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∫r2r2−16​1​dr can be written as k∫cos⁡ϕ 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∫r2r2−16​1​dr=16rr2−16​​+C.

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11.5 Integration by Substitution Questions

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.

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11.5 Integration by Substitution Questions

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