Aspire Faculty ID #18351 · Topic: AMU MCA 2016 · Just now
AMU MCA 2016

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

Solution

$12 - x^2 - 4x = 16 - (x+2)^2$ Let $u = x+2$ Limits: $x=-2 \to u=0$, $x=0 \to u=2$ $\int_{0}^{2} \frac{du}{\sqrt{16 - u^2}}$ $= \sin^{-1}\left(\frac{u}{4}\right)\Big|_0^2$ $= \sin^{-1}\left(\frac{2}{4}\right) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$
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