Question

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

Answer 1

We have,

$\int \frac{\sqrt{a^2-x^2}{x}^2}{d}x$

Let put $x=a\sin \theta ---\left(1\right)$

then $dx=a\cos \theta d\theta$

$=\int \frac{\sqrt{a^2-a^2{\sin }^2\theta }}{a^2{\sin }^2\theta }a\cos \theta d\theta$

$=\int \frac{a\sqrt{1{\sin }^2\theta }}{a^2{\sin }^2\theta }a\cos \theta d\theta$

$=\int \frac{a^2{\cos }^2\theta }{a^2{\sin }^2\theta }d\theta$

$=\int {\cot }^2\theta d\theta$

$=\int ({\csc }^2\theta -1)d\theta$

$=\int {\csc }^2\theta -\int 1d\theta$

$=-\cot \theta -\theta +c$

by equation (1)

$x=a\sin \theta$

$\frac{x}{a}\sin \theta$

$\theta ={\sin }^{-1}\frac{x}{a}$

now,

$-\cot \theta -\theta +c$

$-\cot {\sin }^{-1}\frac{x}{a}-{\sin }^{-1}\frac{x}{a}+c$

Hence this is the answer