int_0^pi sqrt{1+4 sin^2 frac{x}{2}+4 sin frac | vedclass

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Question

(C) Let $I = \int_0^\pi \sqrt{1+4 \sin^2 \frac{x}{2} + 4 \sin \frac{x}{2}} \, dx$. Since $(1 + 2 \sin \frac{x}{2})^2 = 1 + 4 \sin^2 \frac{x}{2} + 4 \s

Vedclass · Accepted Answer

$\pi+4$

Vedclass · Answer

(C) Let $I = \int_0^\pi \sqrt{1+4 \sin^2 \frac{x}{2} + 4 \sin \frac{x}{2}} \, dx$. Since $(1 + 2 \sin \frac{x}{2})^2 = 1 + 4 \sin^2 \frac{x}{2} + 4 \sin \frac{x}{2}$, the integral becomes: $I = \int_0^\pi (1 + 2 \sin \frac{x}{2}) \, dx$. Integrating term by term: $I = [x - 2 \cdot 2 \cos \frac{x}{2}]_0^\pi$. $I = [x - 4 \cos \frac{x}{2}]_0^\pi$. Evaluating at the limits: $I = (\pi - 4 \cos \frac{\pi}{2}) - (0 - 4 \cos 0)$. $I = (\pi - 4(0)) - (0 - 4(1))$. $I = \pi - 0 + 4 = \pi + 4$.

$\int_0^\pi \sqrt{1+4 \sin^2 \frac{x}{2}+4 \sin \frac{x}{2}} \, dx$ is equal to

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