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Mathematics Question on Integrals of Some Particular Functions

The value of the integral ∫−20dx12−x2−4x\int_{-2}^{0}\frac{dx}{\sqrt{12-x^{2}-4x}} is

A

π2\frac{\pi}{2}

B

π6\frac{\pi}{6}

C

π3\frac{\pi}{3}

D

−π6-\frac{\pi}{6}

Answer

π6\frac{\pi}{6}

Explanation

Solution

Let I=∫−20dx12−x2−4xI = \int\limits _{-2}^{0} \frac{dx}{\sqrt{12 - x^2 - 4x}}
=∫−20dx12−(x2+4x+4)+4= \int\limits_{-2}^{0} \frac{dx}{\sqrt{12 - (x^2 + 4x + 4) + 4}}
=∫−20dx16−(x+2)2= \int\limits_{-2}^{0} \frac{dx}{\sqrt{16 - (x + 2)^2}}
=∫−20dx42−(x+2)2= \int\limits_{-2}^{0}\frac{dx}{\sqrt{4^2 - (x+2)^2}}
[sin−1(x+24)]−20\left[sin^{-1}\left(\frac{x+2}{4}\right)\right]_{-2}^{0}
=sin−1(0+24)−sin−1(−2+24)= sin^{-1}\left(\frac{0+2}{4}\right) - sin^{-1}\left(\frac{-2+2}{4}\right)
=sin−1(12)−sin−1(0)= sin^{-1}\left(\frac{1}{2}\right)- sin^{-1}\left(0\right)
=π6−0= \frac{\pi}{6} - 0
=π6= \frac{\pi}{6}