e^{int_0^{pi / 2} sqrt{frac{1-sin 2 x}{1+sin | vedclass

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(D) Let $I = \int_0^{\pi / 2} \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x$. We know that $1-\sin 2x = (\cos x - \sin x)^2$ and $1+\sin 2x = (\cos x + \si

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$2$

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(D) Let $I = \int_0^{\pi / 2} \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x$. We know that $1-\sin 2x = (\cos x - \sin x)^2$ and $1+\sin 2x = (\cos x + \sin x)^2$. So, $\sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} = \left| \frac{\cos x - \sin x}{\cos x + \sin x} ight| = |	an(\frac{\pi}{4} - x)|$. Since $	an(\frac{\pi}{4} - x) \ge 0$ for $x \in [0, \frac{\pi}{4}]$ and $	an(\frac{\pi}{4} - x) $I = \int_0^{\pi / 4} 	an(\frac{\pi}{4} - x) dx + \int_{\pi / 4}^{\pi / 2} -	an(\frac{\pi}{4} - x) dx$. Using $\int 	an(ax+b) dx = -\frac{1}{a} \log|\cos(ax+b)| + C$: $I = [\log|\cos(\frac{\pi}{4} - x)|]_0^{\pi / 4} - [\log|\cos(\frac{\pi}{4} - x)|]_{\pi / 4}^{\pi / 2}$. $I = (\log 1 - \log \frac{1}{\sqrt{2}}) - (\log \frac{1}{\sqrt{2}} - \log 1) = \log \sqrt{2} + \log \sqrt{2} = 2 \log \sqrt{2} = \log 2$. Therefore, the original expression is $e^I = e^{\log 2} = 2$.

$e^{\int_0^{\pi / 2} \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x}=$

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