Question

Evaluate:
$\int \frac{\sqrt{x^2-a^2}}{x}dx$

Answer 1

We have,

$I=\int \frac{\sqrt{x^2-a^2}}{x}dx$

$putx=a\sec \theta$

$dx=a\sec \theta .\tan \theta d\theta$

Therefore,

$I=\int \frac{\sqrt{a^2{\sec }^2\theta -a^2}}{a\sec \theta }.a\sec \theta \tan d\theta =\int a\sqrt{\left({\sec }^2\theta -1\right)}.\tan \theta d\theta =a\int {\tan }^{2}d\theta =a\int \left({\sec }^2\theta -1\right)d\theta =a\tan \theta -a.\theta +c$

On putting the value of $\theta$, we get

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

Hence, this is the answer.