CUET Preparation Today
Evaluate $\int \frac{dx}{\sqrt{16-9x^2}}$. |
$\frac{1}{3} \sin^{-1}\left(\frac{x}{4}\right) + C$ $\frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C$ $\sin^{-1}\left(\frac{3x}{4}\right) + C$ $\frac{1}{4} \sin^{-1}\left(\frac{3x}{4}\right) + C$ |
$\frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C$ |
The correct answer is Option (2) → $\frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C$ Let $I = \int \frac{dx}{\sqrt{16-9x^2}} = \int \frac{dx}{\sqrt{(4)^2 - (3x)^2}} dx$ $= \frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C$ |
