int_0^{2pi } {sqrt {1 + sin frac{x}{2}} ,dx = | vedclass

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(C) We know that $1 + \sin 	heta = (\sin \frac{	heta}{2} + \cos \frac{	heta}{2})^2$. Thus,$\sqrt{1 + \sin \frac{x}{2}} = \sqrt{(\sin \frac{x}{4} +

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) We know that $1 + \sin 	heta = (\sin \frac{	heta}{2} + \cos \frac{	heta}{2})^2$. Thus,$\sqrt{1 + \sin \frac{x}{2}} = \sqrt{(\sin \frac{x}{4} + \cos \frac{x}{4})^2} = |\sin \frac{x}{4} + \cos \frac{x}{4}|$. Since $x \in [0, 2\pi]$,$\frac{x}{4} \in [0, \frac{\pi}{2}]$,where both $\sin \frac{x}{4}$ and $\cos \frac{x}{4}$ are non-negative. Therefore,the integral becomes $\int_0^{2\pi} (\sin \frac{x}{4} + \cos \frac{x}{4}) dx$. Evaluating the integral: $= [-4 \cos \frac{x}{4} + 4 \sin \frac{x}{4}]_0^{2\pi}$ $= (-4 \cos \frac{\pi}{2} + 4 \sin \frac{\pi}{2}) - (-4 \cos 0 + 4 \sin 0)$ $= (-4(0) + 4(1)) - (-4(1) + 4(0))$ $= 4 - (-4) = 8$.

$\int_0^{2\pi } {\sqrt {1 + \sin \frac{x}{2}} \,dx = } $

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