question_answer

The integral \[\int\limits_{0}^{\pi }{\sqrt{1+4{{\sin }^{2}}\frac{x}{2}-4\sin \frac{x}{2}-4\sin \frac{x}{2}dx}}\]equals:

A) \[\pi -4\]

B)                                \[\frac{2\pi }{3}-4-4\sqrt{3}\]

C) \[4\sqrt{3}-4\]   

D)        \[4\sqrt{3}-4-\frac{\pi }{3}\]

Correct Answer: D

Solution :

\[I=\int\limits_{0}^{\pi }{\sqrt{1+4{{\sin }^{2}}\frac{x}{2}-4\sin \frac{x}{2}dx}}\]

\[=\int\limits_{0}^{\pi }{\left| 1-2\sin \frac{x}{2} \right|dx}\]

\[=\int\limits_{0}^{\pi /3}{\left( 1-2\sin \frac{x}{2} \right)dx\int\limits_{\pi /3}^{\pi }{\left( 2\sin \frac{x}{2}-1 \right)dx}}\]

\[=\left. \left( x+4\cos e\frac{x}{2} \right) \right|_{0}^{\pi /3}+\left. \left( -\,4\cos e\frac{\pi }{2}-x \right) \right|_{\pi /3}^{\pi }\]

\[=\frac{\pi }{3}+8.\frac{\sqrt{3}}{2}-4\]

\[=4\sqrt{3}-4-\frac{\pi }{3}.\]