Question

Evaluate the following integrals:

\[\int\frac{x^7}{\left( a^2 - x^2 \right)^5}dx\]

Answer 1

\[\text{Let I} = \int\frac{x^7}{\left( a^2 - x^2 \right)^5}dx\]

\[ \text{Let x} = a \sin\theta\]

\[ \text{On differentiating both sides, we get}\]

`  dx =  a  cos  θ  dθ `

\[ \therefore I = \int\frac{a^8 \sin^7 \theta \cos\theta}{\left( a^2 - a^2 \sin^2 \theta \right)^5}d\theta\]

\[ = \int\frac{a^8 \sin^7 \theta \cos\theta}{a^{10} \left( 1 - \sin^2 \theta \right)^5}d\theta\]

\[ = \int\frac{\sin^7 \theta}{a^2 \cos^9 \theta}d\theta\]

\[ = \frac{1}{a^2}\int \tan^7 \theta \sec^2 \theta d\theta\]

\[\]

\[ \text{Let} \tan\theta = t\]

` " On differentiating both sides, we get" `

`sec^2 θ  dθ  = dt`

\[ \therefore I = \frac{1}{a^2}\int t^7 dt\]

\[ = \frac{1}{a^2}\frac{t^8}{8} + c\]

\[ = \frac{1}{8 a^2}\left( \tan^8 \theta \right) + c\]

\[ = \frac{1}{8 a^2} \left( \tan\left( \sin^{- 1} \frac{x}{a} \right) \right)^8 + c\]

\[ = \frac{1}{8 a^2} \left( \tan\left( \tan^{- 1} \frac{x}{\sqrt{a^2 - x^2}} \right) \right)^8 + c\]

\[ = \frac{1}{8 a^2} \left( \frac{x}{\sqrt{a^2 - x^2}} \right)^8 + c\]

\[ = \frac{1}{8 a^2}\frac{x^8}{\left( a^2 - x^2 \right)^4} + c\]

\[Hence, \int\frac{x^7}{\left( a^2 - x^2 \right)^5}dx = \frac{1}{8 a^2}\frac{x^8}{\left( a^2 - x^2 \right)^4} + c\]