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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