Question

Solve: $I=\stackrel{\pi /4}{\underset{0}{\int }}\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}dx$

Answer 1

${\int }_0^{\frac{\pi }{4}}\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}dx$

$={\int }_0^{\frac{\pi }{2}}\sqrt{\frac{{\cos }^{2}x+{\sin }^{2}x-2\sin x\cos x}{{\cos }^{2}x+{\sin }^{2}x+2\sin x\cos x}}dx\left(\begin{array}{c}\sin 2\theta =2\sin \theta \cos \theta \\ {\cos }^2\theta +{\sin }^2\theta =1\end{array}\right)$

$={\int }_0^{\frac{\pi }{2}}\sqrt{\frac{(\cos x.\sin x{)}^2}{(\cos x+\sin x{)}^2}}dx$

$={\int }_0^{\frac{\pi }{2}}\frac{\cos x.\sin x}{\cos x+\sin x}dx$

${\int }_0^{\frac{\pi }{2}}\frac{1-\tan x}{1+\tan x}dx$ (Taking $\cos x$ common from numerator and denominator )

$={\int }_0^{\frac{\pi }{2}}\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x}$ $(\because \tan \frac{\pi }{4}=1)$

$={\int }_0^{\frac{\pi }{4}}\tan (\frac{\pi }{4}-x)dx$

$=-{\sec }^2(\frac{\pi }{4}-x){|}_0^{\frac{\pi }{4}}$

$-({\sec }^2\theta -{\sec }^2\frac{\pi }{4})$

$=-(1-2)$

$=-(-1)=1$

$\therefore {\int }_0^{\frac{\pi }{4}}\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}dx=1$