# Evaluate the Following Integrals: ∫ X 7 ( a 2 − X 2 ) 5 D X - Mathematics

Sum

Evaluate the following integrals:

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

#### Solution

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

$\text{Let x} = a \sin\theta$

$\text{On differentiating both sides, we get}$

  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

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

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

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

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

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

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

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

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

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.13 | Q 2 | Page 79