int sqrt{1 + sin frac{x}{2}} , dx = | vedclass

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(C) We know that $1 = \sin^2 \frac{x}{4} + \cos^2 \frac{x}{4}$ and $\sin \frac{x}{2} = 2 \sin \frac{x}{4} \cos \frac{x}{4}$.Substituting these into th

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$4\left( \sin \frac{x}{4} - \cos \frac{x}{4} \right) + c$

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(C) We know that $1 = \sin^2 \frac{x}{4} + \cos^2 \frac{x}{4}$ and $\sin \frac{x}{2} = 2 \sin \frac{x}{4} \cos \frac{x}{4}$.Substituting these into the integral,we get:$\int \sqrt{\sin^2 \frac{x}{4} + \cos^2 \frac{x}{4} + 2 \sin \frac{x}{4} \cos \frac{x}{4}} \, dx$$= \int \sqrt{(\sin \frac{x}{4} + \cos \frac{x}{4})^2} \, dx$$= \int (\sin \frac{x}{4} + \cos \frac{x}{4}) \, dx$$= -4 \cos \frac{x}{4} + 4 \sin \frac{x}{4} + c$$= 4(\sin \frac{x}{4} - \cos \frac{x}{4}) + c$.

$\int \sqrt{1 + \sin \frac{x}{2}} \, dx = $

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