If frac{5 pi}{4}

Question

(B) We have the integrand $\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$.Using the identities $1-\sin 2x = (\cos x - \sin x)^2$ and $1+\sin 2x = (\cos x + \sin

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$-\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+c$

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(B) We have the integrand $\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$.Using the identities $1-\sin 2x = (\cos x - \sin x)^2$ and $1+\sin 2x = (\cos x + \sin x)^2$, we get:$\sqrt{\frac{(\cos x - \sin x)^2}{(\cos x + \sin x)^2}} = \left| \frac{\cos x - \sin x}{\cos x + \sin x} ight| = \left| \frac{1 - 	an x}{1 + 	an x} ight| = |	an(\frac{\pi}{4} - x)|$.Given $\frac{5\pi}{4} In this interval, $	an(\frac{\pi}{4} - x)$ is positive, so $|	an(\frac{\pi}{4} - x)| = 	an(\frac{\pi}{4} - x)$.Now, $\int 	an(\frac{\pi}{4} - x) dx = -\ln|\sec(\frac{\pi}{4} - x)| + c$.

If $\frac{5 \pi}{4} < x < \frac{7 \pi}{4}$, then $\int \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x=$

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