JEE Main 2014 — Definite Integration Question with Solution

$\underset{0}{\overset{\pi }{\int }}\sqrt{1+\mathrm{4 }{\text{sin}}^2\frac{x}{2}-\mathrm{4 }\text{sin}\frac{x}{2}}dx$

$=\underset{0}{\overset{\pi }{\int }}\sqrt{{\left(\mathrm{2 }\text{sin}\frac{x}{2}-1\right)}^2}dx$

$=\underset{0}{\overset{\pi }{\int }}\left|\mathrm{2 }\text{sin}\frac{x}{2}-1\right|dx$

$=2\underset{0}{\overset{\pi }{\int }}\left|\text{sin}\frac{x}{2}-\frac{1}{2}\right|dx$

$=\mathrm{2 }\left[\underset{0}{\overset{\pi /3}{\int }}\left|\text{sin}\frac{x}{2}-\frac{1}{2}\right|dx+\underset{\pi /3}{\overset{\pi }{\int }}\left|\text{sin}\frac{x}{2}-\frac{1}{2}\right|dx\right]$

$=\mathrm{2 }\left[\underset{0}{\overset{\pi /3}{\int }}-\left(\text{sin}\frac{x}{2}-\frac{1}{2}\right)dx+\underset{\pi /3}{\overset{\pi }{\int }}\left(\text{sin}\frac{x}{2}-\frac{1}{2}\right)dx\right]$

$=\mathrm{2 }\left[{\left[2\text{cos}\frac{x}{2}+\frac{x}{2}\right]}_0^{\pi /3}+{\left[-2\text{cos}\frac{x}{2}-\frac{x}{2}\right]}_{\pi /3}^{\pi }\right]$

$=\mathrm{2 }\left[\left[2\left(\frac{\sqrt{3}}{2}\right)+\frac{\pi }{6}-2\right]+\left[-\frac{\pi }{2}+2\left(\frac{\sqrt{3}}{2}\right)+\frac{\pi }{6}\right]\right]$

$=\mathrm{2 }\left[\sqrt{3}+\frac{\pi }{6}-2-\frac{\pi }{2}+\sqrt{3}+\frac{\pi }{6}\right]$

$=\mathrm{2 }\left[2\sqrt{3}-\frac{\pi }{6}-2\right]$

$=4\sqrt{3}-\frac{\pi }{3}-4$.