Question

$\int_{-1/2}^{1/2} \frac{dx}{\sqrt{1-x^2}}$ is equal to

• A. $\frac{\pi}{3}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{2}$
• D. $0$

Hint:

$\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C$.

Answer 1

The Correct Option is A

Step 1: Basic Principle
$\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C$.
Step 2: Solution Procedure:
$\int_{-1/2}^{1/2} \frac{dx}{\sqrt{1-x^2}} = [\sin^{-1}x]_{-1/2}^{1/2} = \sin^{-1}(1/2) - \sin^{-1}(-1/2) = \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{3}$.
Step 3: Required Answer:
The integral equals $\frac{\pi}{3}$.