Question

Integrate the function: $\frac{1}{\sqrt{9-25x^2}}$

Answer 1

$\int \frac{1}{\sqrt{9-25x^2}}dx$

$=\int \frac{1}{\sqrt{25(\frac{9}{25}-x^2)}}dx$

$=\frac{1}{5}\int \frac{1}{\sqrt{\frac{9}{25}-x^2}}dx$

$=\frac{1}{5}\int \frac{1}{\sqrt{{\left(\frac{3}{5}\right)}^2-x^2}}dx$

$=\frac{1}{5}{\sin }^{-1}\left(\frac{\frac{x}{3}}{5}\right)+C$

$[\because \int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\left(\frac{x}{a}\right)+C]$

$=\frac{1}{5}{\sin }^{-1}\left(\frac{5x}{3}\right)+C$

Where $C$ is constant of integration.