QJEEM14Ma239c: JEE Main 2014 Mathematics - Class 12 JEEM14Ma239c Solution: The integral equals:

Question

Verified Solution • 2025–26 Session moderate

The integral

\int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}{\mkern 1mu} } } dx

equals:

Accepted Answer

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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