Question:
$\int \frac{d x}{\sqrt{16-9 x^{2}}}$
Solution:
Let, $\mathrm{I}=\int \frac{d x}{\sqrt{16-9 x^{2}}}$
$=\frac{1}{3} \int \frac{d x}{\sqrt{\frac{16}{9}-x^{2}}}=\frac{1}{3} \int \frac{d x}{\sqrt{\left(\frac{4}{2}\right)^{2}-x^{2}}}$
$=\frac{1}{3} \sin ^{-1} \frac{x}{4 / 3}+C\left[\because \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}+C\right]$
$=\frac{1}{3} \sin ^{-1} \frac{3 x}{4}+C$
Therefore, $\mathrm{I}=\frac{1}{3} \sin ^{-1} \frac{3 x}{4}+C$.