JEE Main · 2014 · 06 Apr 2014 · Offline · Definite Integrals

The quantity under the root is a perfect square: $1+4{\sin }^2\frac{x}{2}-4\sin \frac{x}{2}={\left(1-2\sin \frac{x}{2}\right)}^2$, so the integrand is $\left|1-2\sin \frac{x}{2}\right|$. This changes sign where $\sin \frac{x}{2}=\frac{1}{2}$, that is at $x=\frac{\pi }{3}$: it is non-negative on $\left[0,\frac{\pi }{3}\right]$ and negative on $\left(\frac{\pi }{3},\pi \right]$. Splitting there, $I={\int }_0^{\pi /3}\left(1-2\sin \frac{x}{2}\right)dx+{\int }_{\pi /3}^{\pi }\left(2\sin \frac{x}{2}-1\right)dx$, and with $\int \left(1-2\sin \frac{x}{2}\right)dx=x+4\cos \frac{x}{2}$ the two parts combine to $4\sqrt{3}-4-\frac{\pi }{3}$.