Q71 / 170 · Manipal (MET) 2017 · Mathematics
$\int_{-1 / 2}^{1 / 2} \frac{d x}{\left(1-x^2\right)^{1 / 2}}$ is equal to
- Correct answerA$\frac{\pi}{3}$
- B$\frac{\pi}{4}$
- C$\frac{\pi}{2}$
- D$0$
Solution
Let $I=\int_{-1 / 2}^{1 / 2} \frac{d x}{\left(1-x^2\right)^{1 / 2}}$
Again, let $f(x)=\frac{1}{\left(1-x^2\right)^{1 / 2}}$
$\begin{gathered}
f(-x)=\frac{1}{\left(1-x^2\right)^{1 / 2}} = f(x) \\
\therefore I=2 \int_0^{1 / 2} \frac{d x}{\left(1-x^2\right)^{1 / 2}}=\left[2 \sin ^{-1} \frac{x}{1}\right]_0^{1 / 2} \\
\end{gathered}$
$=2 \sin ^{-1} \frac{1}{2}=2 \times \frac{\pi}{6}=\frac{\pi}{3}$
Hence, the correct answer is option: 'A'.