The value of \(\int_{0}^{\frac{1}{2}}\displaystyle\frac{dx}{\sqrt{1-x^{2}}}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\pi}{6}\)
Since \(\displaystyle\int\frac{dx}{\sqrt{1-x^{2}}} = \sin^{-1}x + C\),
\( \displaystyle \int_{0}^{1/2}\frac{dx}{\sqrt{1-x^{2}}} = \left[\sin^{-1}x\right]_{0}^{1/2} \)
\( \displaystyle = \sin^{-1}\frac{1}{2} - \sin^{-1}0 \)
\( \displaystyle = \frac{\pi}{6} - 0 = \frac{\pi}{6} \)