mathop smallint limits_0^pi sqrt {1 + 4{{sin | vedclass

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(B) The given integral is $I = \int_0^\pi \sqrt{1 + 4\sin^2\frac{x}{2} - 4\sin\frac{x}{2}} \; dx$.This can be written as $I = \int_0^\pi \sqrt{(1 - 2\

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

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(B) The given integral is $I = \int_0^\pi \sqrt{1 + 4\sin^2\frac{x}{2} - 4\sin\frac{x}{2}} \; dx$.This can be written as $I = \int_0^\pi \sqrt{(1 - 2\sin\frac{x}{2})^2} \; dx = \int_0^\pi |1 - 2\sin\frac{x}{2}| \; dx$.We know $1 - 2\sin\frac{x}{2} = 0$ when $\sin\frac{x}{2} = \frac{1}{2}$,which means $\frac{x}{2} = \frac{\pi}{6}$,so $x = \frac{\pi}{3}$.For $0 \le x  0$. For $\frac{\pi}{3} Thus,$I = \int_0^{\pi/3} (1 - 2\sin\frac{x}{2}) \; dx + \int_{\pi/3}^\pi -(1 - 2\sin\frac{x}{2}) \; dx$.$I = [x + 4\cos\frac{x}{2}]_0^{\pi/3} - [x + 4\cos\frac{x}{2}]_{\pi/3}^\pi$.$I = ((\frac{\pi}{3} + 4\cos\frac{\pi}{6}) - (0 + 4\cos 0)) - ((\pi + 4\cos\frac{\pi}{2}) - (\frac{\pi}{3} + 4\cos\frac{\pi}{6}))$.$I = (\frac{\pi}{3} + 4(\frac{\sqrt{3}}{2}) - 4) - ((\pi + 0) - (\frac{\pi}{3} + 4(\frac{\sqrt{3}}{2})))$.$I = (\frac{\pi}{3} + 2\sqrt{3} - 4) - (\pi - \frac{\pi}{3} - 2\sqrt{3}) = \frac{\pi}{3} + 2\sqrt{3} - 4 - \frac{2\pi}{3} + 2\sqrt{3} = 4\sqrt{3} - 4 - \frac{\pi}{3}$.

$\mathop \smallint \limits_0^\pi \sqrt {1 + 4{{\sin }^2}\frac{x}{2} - 4\sin \frac{x}{2}} \;dx = $

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