Question
The integral $$\int\limits_0^\pi {\sqrt {1 + 4\,{{\sin }^2}\frac{x}{2} - 4\,\sin \frac{x}{2}} } \,dx$$ equals:
A.
$$4\sqrt 3 - 4$$
B.
$$4\sqrt 3 - 4 - \frac{\pi }{3}$$
C.
$$\pi - 4$$
D.
$$\frac{{2\pi }}{3} - 4 - 4\sqrt 3 $$
Answer :
$$4\sqrt 3 - 4 - \frac{\pi }{3}$$
Solution :
$$\eqalign{
& {\text{Let }}I = \int\limits_0^\pi {\sqrt {1 + 4\,{{\sin }^2}\frac{x}{2} - 4\,\sin \frac{x}{2}} } \,dx \cr
& = \int\limits_0^\pi {\left| {2\sin \,\frac{x}{2} - 1} \right|dx} \cr
& = \int\limits_0^{\frac{\pi }{3}} {\left( {1 - 2\,\sin \,\frac{x}{2}} \right)dx} + \int\limits_{\frac{\pi }{3}}^\pi {\left( {2\,\sin \,\frac{x}{2} - 1} \right)} \,dx \cr
& \left[ {\because \sin \frac{x}{2} = \frac{1}{2} \Rightarrow \frac{x}{2} = \frac{\pi }{6} \Rightarrow x = \frac{\pi }{3},\,\frac{x}{2} = \frac{{5\pi }}{6} \Rightarrow x = \frac{{5\pi }}{3}} \right] \cr
& = \left[ {x + 4\,\cos \frac{x}{2}} \right]_0^{\frac{\pi }{3}} + \left[ { - 4\,\cos \frac{x}{2} - x} \right]_{\frac{\pi }{3}}^\pi \cr
& = \frac{\pi }{3} + 4\frac{{\sqrt 3 }}{2} - 4 + \left( {0 - \pi + 4\frac{{\sqrt 3 }}{2} + \frac{\pi }{3}} \right) \cr
& = 4\sqrt 3 - 4 - \frac{\pi }{3} \cr} $$