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

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(B) Given $I = \int \sqrt{\frac{2}{1+\sin x}} dx$. Using $\sin x = \cos(\frac{\pi}{2} - x)$,we have $1 + \sin x = 1 + \cos(\frac{\pi}{2} - x) = 2 \cos

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$\frac{1}{\sqrt{3+2 \sqrt{2}}}$

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(B) Given $I = \int \sqrt{\frac{2}{1+\sin x}} dx$. Using $\sin x = \cos(\frac{\pi}{2} - x)$,we have $1 + \sin x = 1 + \cos(\frac{\pi}{2} - x) = 2 \cos^2(\frac{\pi}{4} - \frac{x}{2})$. So,$\sqrt{\frac{2}{1+\sin x}} = \sqrt{\frac{2}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})}} = \frac{1}{|\cos(\frac{\pi}{4} - \frac{x}{2})|} = \sec(\frac{\pi}{4} - \frac{x}{2})$. Since $0 \leq x \leq \frac{\pi}{2}$,$\frac{\pi}{4} - \frac{x}{2}$ lies in $[0, \frac{\pi}{4}]$,where $\cos$ is positive. Thus,$I = \int \sec(\frac{\pi}{4} - \frac{x}{2}) dx = -2 \log |\sec(\frac{\pi}{4} - \frac{x}{2}) + 	an(\frac{\pi}{4} - \frac{x}{2})| + C$. Using $\sec 	heta + 	an 	heta = 	an(\frac{\pi}{4} + \frac{	heta}{2})$,we can write this as $2 \log |\sec(\frac{x}{2} - \frac{\pi}{4}) + 	an(\frac{x}{2} - \frac{\pi}{4})| + C$. Comparing with $2 \log |A(x) - B(x)|$,we identify $B(x) = -	an(\frac{x}{2} - \frac{\pi}{4}) = 	an(\frac{\pi}{4} - \frac{x}{2})$. Then $B(\frac{\pi}{4}) = 	an(\frac{\pi}{4} - \frac{\pi}{8}) = 	an(\frac{\pi}{8})$. Since $	an(\frac{\pi}{4}) = \frac{2 	an(\pi/8)}{1 - 	an^2(\pi/8)} = 1$,let $y = 	an(\frac{\pi}{8})$. Then $2y = 1 - y^2 \Rightarrow y^2 + 2y - 1 = 0$. Solving for $y > 0$,$y = \frac{-2 + \sqrt{4 + 4}}{2} = \sqrt{2} - 1$. Thus $B(\frac{\pi}{4}) = \sqrt{2} - 1 = \frac{1}{\sqrt{2} + 1} = \frac{1}{\sqrt{(\sqrt{2} + 1)^2}} = \frac{1}{\sqrt{3 + 2\sqrt{2}}}$.

If $\int \sqrt{\frac{2}{1+\sin x}} dx = 2 \log |A(x) - B(x)| + C$ and $0 \leq x \leq \frac{\pi}{2}$,then $B(\frac{\pi}{4}) = $

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