Question

The value of the integral $ \int_{-2}^{0}\frac{dx}{\sqrt{12-x^{2}-4x}} $ is

• A. $ \frac{\pi}{2} $
• B. $ \frac{\pi}{6} $
• C. $ \frac{\pi}{3} $
• D. $ -\frac{\pi}{6} $

Answer 1

The Correct Option is B

Let $I = \int\limits _{-2}^{0} \frac{dx}{\sqrt{12 - x^2 - 4x}}$
$ = \int\limits_{-2}^{0} \frac{dx}{\sqrt{12 - (x^2 + 4x + 4) + 4}}$
$= \int\limits_{-2}^{0} \frac{dx}{\sqrt{16 - (x + 2)^2}}$
$ = \int\limits_{-2}^{0}\frac{dx}{\sqrt{4^2 - (x+2)^2}}$
$\left[sin^{-1}\left(\frac{x+2}{4}\right)\right]_{-2}^{0} $
$= sin^{-1}\left(\frac{0+2}{4}\right) - sin^{-1}\left(\frac{-2+2}{4}\right) $
$= sin^{-1}\left(\frac{1}{2}\right)- sin^{-1}\left(0\right)$
$= \frac{\pi}{6} - 0$
$ = \frac{\pi}{6}$