The value of $\int\limits_{0}^{3} \frac{dx}{\sqrt{9-x^2}}$ is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{\pi}{2}$
$\int\limits_{0}^{3} \frac{dx}{\sqrt{9-x^2}}$ $\left[ ∵ \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \frac{x}{a} + C \right]$
$I = \int\limits_{0}^{3} \frac{dx}{\sqrt{3^2 - x^2}}$
$I = \left[ \sin^{-1} \frac{x}{3} \right]_0^3 = \left[ \sin^{-1} \frac{3}{3} - \sin^{-1} \frac{0}{3} \right]$
$= \frac{\pi}{2}$