Question

Evaluate the following integrals:

$\int \frac{x^7}{{\left(a^2-x^2\right)}^{{\displaystyle 5}}}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^7}{{\left(a^2-x^2\right)}^{{\displaystyle 5}}}\mathrm{d}x \\ \mathrm{Let}x=a\sin \theta \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \mathrm{d}x=a\cos \theta \mathrm{d}\theta \\ \therefore I=\int \frac{a^8{\sin }^7\theta \cos \theta }{{\left(a^2-a^2{\sin }^2\theta \right)}^{{\displaystyle 5}}}\mathrm{d}\theta \\ =\int \frac{a^8{\sin }^7\theta \cos \theta }{a^{10}{\left(1-{\sin }^2\theta \right)}^{{\displaystyle 5}}}\mathrm{d}\theta \\ =\int \frac{{\sin }^7\theta }{a^2{\cos }^9\theta }\mathrm{d}\theta \\ =\frac{1}{a^2}\int {\tan }^7\theta {\sec }^2\theta \mathrm{d}\theta \\ \mathrm{Let}\tan \theta =t \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ {\sec }^2\theta \mathrm{d}\theta =\mathrm{d}t \\ \therefore I=\frac{1}{a^2}\int t^7\mathrm{d}t \\ =\frac{1}{a^2}\frac{t^8}{8}+c \\ =\frac{1}{8a^2}\left({\tan }^8\theta \right)+c \\ =\frac{1}{8a^2}{\left(\tan \left({\sin }^{-1}\frac{x}{a}\right)\right)}^8+c \\ =\frac{1}{8a^2}{\left(\tan \left({\tan }^{-1}\frac{x}{\sqrt{a^2-x^2}}\right)\right)}^8+c \\ =\frac{1}{8a^2}{\left(\frac{x}{\sqrt{a^2-x^2}}\right)}^8+c \\ =\frac{1}{8a^2}\frac{x^8}{{\left(a^2-x^2\right)}^4}+c \\ \mathrm{Hence},\int \frac{x^7}{{\left(a^2-x^2\right)}^{{\displaystyle 5}}}\mathrm{d}x=\frac{1}{8a^2}\frac{x^8}{{\left(a^2-x^2\right)}^4}+c \end{gathered}$$