The value of $\int\limits_{0}^{3} \frac{dx}{\sqrt{9-x^2}}$ is:

$\frac{\pi}{2}$

The correct answer is Option (3) → $\frac{\pi}{2}$

$\int\limits_{0}^{3} \frac{dx}{\sqrt{9-x^2}}$  $\left[ ∵ \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \frac{x}{a} + C \right]$

$I = \int\limits_{0}^{3} \frac{dx}{\sqrt{3^2 - x^2}}$

$I = \left[ \sin^{-1} \frac{x}{3} \right]_0^3 = \left[ \sin^{-1} \frac{3}{3} - \sin^{-1} \frac{0}{3} \right]$

$= \frac{\pi}{2}$