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The integral $$\int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}{\mkern 1mu} } } dx$$ equals:

JEE · Math · previous-year question

  1. A.$$4\sqrt 3 - 4$$
  2. B.$$4\sqrt 3 - 4 - {\pi \over 3}$$correct
  3. C.$$\pi - 4$$
  4. D.$${{2\pi } \over 3} - 4 - 4\sqrt 3 $$

Answer

B. $$4\sqrt 3 - 4 - {\pi \over 3}$$

Explanation

Let $$I = \int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}} } dx$$ $$ = \int\limits_0^\pi {\left| {2\sin {x \over 2} - 1} \right|} dx$$ $$ = \int\limits_0^{\pi /3} {\left( {1 - 2\sin {x \over 2}} \right)} dx + \int\limits_{\pi /3}^\pi {\left( {2\sin {x \over 2} - 1} \right)} dx$$ $$\left[ {} \right.$$ As $$\sin {x \over 2} = {1 \over 2} \Rightarrow {x \over 2} = {\pi \over 6} \Rightarrow x = {\pi \over 3},$$ $${x \over 2}$$ $$\,\,\,\,\,\,\,\,\,\,\, = {{5\pi } \over 6} \Rightarrow x = {{5\pi } \over 3}$$ $$\left. {} \right]$$ $$ = \left[ {x + 4\cos {x \over 2}} \right]_0^{\pi /3} + \left[ { - 4\cos {x \over 2} - x} \right]_{\pi /3}^\pi $$ $$ = {\pi \over 3} + 4{{\sqrt 2 } \over 2} - 4 + \left( {0 - \pi + 4{{\sqrt 3 } \over 2} + {\pi \over 3}} \right)$$ $$ = 4\sqrt 3 - 4 - {\pi \over 3}$$

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