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

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(D) We are given the integral $I = \int \sqrt{1 + \sin \frac{x}{2}} dx$. Using the identity $1 + \sin 	heta = \sin^2 \frac{	heta}{2} + \cos^2 \frac{

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$4\,\sqrt{2}$

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(D) We are given the integral $I = \int \sqrt{1 + \sin \frac{x}{2}} dx$. Using the identity $1 + \sin 	heta = \sin^2 \frac{	heta}{2} + \cos^2 \frac{	heta}{2} + 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2} = (\sin \frac{	heta}{2} + \cos \frac{	heta}{2})^2$,we have: $1 + \sin \frac{x}{2} = (\sin \frac{x}{4} + \cos \frac{x}{4})^2$. Thus,$\sqrt{1 + \sin \frac{x}{2}} = |\sin \frac{x}{4} + \cos \frac{x}{4}|$. Assuming the interval where $\sin \frac{x}{4} + \cos \frac{x}{4} > 0$,we have: $I = \int (\sin \frac{x}{4} + \cos \frac{x}{4}) dx$. Integrating term by term: $I = -4 \cos \frac{x}{4} + 4 \sin \frac{x}{4} + C$. We can rewrite this as: $I = 4 (\sin \frac{x}{4} - \cos \frac{x}{4}) + C$. Using the identity $\sin 	heta - \cos 	heta = \sqrt{2} \sin(	heta - \frac{\pi}{4})$,we get: $I = 4 \sqrt{2} \sin(\frac{x}{4} - \frac{\pi}{4}) + C$. Comparing this with the given form $A \sin(\frac{x}{4} - \frac{\pi}{4})$,we find $A = 4\sqrt{2}$.

If $\int {\sqrt {1 + \sin \frac{x}{2}} } dx = A\, \sin\, \left( {\frac{x}{4} - \frac{\pi }{4}} \right) + C$,then the value of $A$ is:

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