Evaluate $\int \frac{dx}{\sqrt{16-9x^2}}$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C$
Let $I = \int \frac{dx}{\sqrt{16-9x^2}} = \int \frac{dx}{\sqrt{(4)^2 - (3x)^2}} dx$
$= \frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C$