JEE-Mains 2014 (15/30)

$\int_0^\pi \sqrt{1+4\sin^2{\frac{x}{2}}-4\sin{\frac{x}{2}}}dx=\int_0^\pi \sqrt{\Big(2\sin{\frac{x}{2}}-1\Big)^2}dx=\int_0^\pi |2\sin{\frac{x}{2}}-1|dx=\Big(\int_{\pi/3}^\pi-\int_0^{\pi/3}\Big)\Big(\Big(2\sin{\frac{x}{2}}-1\Big)dx\Big)$ $=\Big(-4\cos{\frac{x}{2}}-x\Big)\Big(\Big|_{\pi/3}^\pi-\Big|_0^{\pi/3}\Big)=-4\cos{\frac{\pi}{2}}-\pi+4\cos{\frac{\pi}{3}}+\frac{\pi}{3}-\Big(-4\cos{\frac{\pi}{3}}-\frac{\pi}{3}+4\cos{0}+0\Big)=4\sqrt{3}-4-\frac{\pi}{3}$

Similar solution with @James Wilson 's

$\begin{aligned} I & = \int_0^\pi \sqrt{1+4\sin^2 \frac x2 - 4\sin \frac x2}\ dx \\ & = \int_0^\pi \sqrt{\left(2\sin \frac x2 - 1\right)^2} dx & \small \color{#3D99F6} \text{Note that }\sqrt{\left(2\sin \frac x2 - 1\right)^2} \ge 0 \text{ for }x \in [0, \pi] \\ & = \int_0^\pi \left|2\sin \frac x2 - 1\right| dx & \small \color{#3D99F6} \text{but }2\sin \frac x2 - 1 < 0 \text{ for }x \in \left[0, \frac \pi 3 \right) \\ & = \int_0^\frac \pi 3 \left(1-2\sin \frac x2\right) dx + \int_\frac \pi 3^\pi \left(2\sin \frac x2-1\right) dx \\ & = x + 4\cos \frac x2 \ \bigg|_0^\frac \pi 3 - 4\cos \frac x2 - x\ \bigg|^\pi_\frac \pi 3 \\ & = \frac \pi 3 + 2\sqrt 3 - 0 - 4 - 0 - \pi + 2\sqrt 3 + \frac \pi 3 \\ & = \boxed{4\sqrt 3-4-\dfrac \pi 3} \end{aligned}$