Question

Solve
$\int \frac{1}{(a^2-x^2{)}^{3/2}}dx$

Answer 1

Given

$\int \frac{1}{(a^2-x^2{)}^{\frac{3}{2}}}$

Substitute $x=a\sin u$, $u={\sin }^{-1}\left(\frac{x}{a}\right)$, $dx=a\cos udu$

$=\int \frac{a\cos u}{(a^2-a^2{\sin }^{2}u{)}^{\frac{1}{2}}}du$

Now, $a^2-a^2{\sin }^{2}u=a^2(1-{\sin }^{2}u)=a^2{\cos }^{2}u$

$=\frac{1}{a^2}\int \frac{1}{{\cos }^{2}u}du$

$=\frac{1}{a^2}\int {\sec }^{2}udu$

$=\frac{\tan u}{a^2}$

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

$=\frac{x}{a^3\sqrt{1-\frac{x^2}{a^2}}}$

$=\frac{x}{a^3\sqrt{\frac{a^2-x^2}{a^2}}}$

$=\frac{ax}{a^3\sqrt{a^2-x^2}}$

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