The value of _0^3 d x 9-x^2 is: - Vidyadip

MCQ

The value of $\int_0^3 \frac{d x}{\sqrt{9-x^2}}$ is:

A
$\frac{\pi}{6}$
B
$\frac{\pi}{4}$
C
$\frac{\pi}{2}$
D
$\frac{\pi}{18}$

✓

Answer

We have, $\int_0^3 \frac{d x}{\sqrt{9-x^2}}=\left[\sin ^{-1} \frac{x}{3}\right]_0^3=\sin ^{-1} 1-\sin ^{-1} 0=\frac{\pi}{2}$

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