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$\int_0^\pi \sqrt{1+4 \sin ^2 \frac{x}{2}+4 \sin \frac{x}{2}} d x$ is equal to
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$\pi+4$
$I=\int_0^\pi \sqrt{1+4 \sin ^2(x / 2)+4 \sin (x / 2)} d x$
$\begin{aligned} & \because \quad\left(1+2 \sin \frac{x}{2}\right)^2=1+4 \sin ^2 x / 2+4 \sin (x / 2) \\ & \therefore \quad I=\int_0^\pi(1+2 \sin x / 2) d x=\left[x-\frac{2}{1 / 2} \cos \frac{x}{2}\right]_0^\pi \\ & \quad=[x-4 \cos (x / 2)]_0^\pi=[\pi-4.0-0+4]=\pi+4\end{aligned}$
$\begin{aligned} & \because \quad\left(1+2 \sin \frac{x}{2}\right)^2=1+4 \sin ^2 x / 2+4 \sin (x / 2) \\ & \therefore \quad I=\int_0^\pi(1+2 \sin x / 2) d x=\left[x-\frac{2}{1 / 2} \cos \frac{x}{2}\right]_0^\pi \\ & \quad=[x-4 \cos (x / 2)]_0^\pi=[\pi-4.0-0+4]=\pi+4\end{aligned}$
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