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Q. The integral $ \int\limits^{\pi}_0 \sqrt { 1 + 4\, \sin^2 \frac{x}{2} - 4 \, \sin \frac{x}{2} \, dx }$ is equal to

JEE MainJEE Main 2014Integrals

Solution:

PLAN Use the formula, $|x-a|= \begin{cases} x-a, & x \geq a \\ -(x-a), & x < a\end{cases}$
to break given integral in two parts and then integrate separately.
$\int\limits_{0}^{\pi} \sqrt{\left(1-2 \sin \frac{x}{2}\right)^{2}} d x=\int\limits_{0}^{\pi}\left|1-2 \sin \frac{x}{2}\right| d x $
$=\int\limits_{0}^{\frac{\pi}{3}}\left(1-2 \sin \frac{x}{2}\right) d x-\int\limits_{\frac{\pi}{3}}^{\pi}\left(1-2 \sin \frac{x}{2}\right) d x$
$=\left(x+4 \cos \frac{x}{2}\right)_{0}^{\frac{\pi}{3}}-\left(x+4 \cos \frac{x}{2}\right)_{\frac{\pi}{3}}^{\pi}$
$=4 \sqrt{3}-4-\frac{\pi}{3}$