Question

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

Answer 1

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

Let $5x=t$

$5dx=dt$

Therefore,

$I=\frac{1}{5}\int \frac{dt}{\sqrt{3^2-t^2}}$

$I=\frac{1}{5}{\sin }^{-1}\frac{t}{3}+C$ ............ $\because \int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\left(\frac{x}{a}\right)+c$

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