Question

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

• A. $\log 2$
• B. $-\log \sqrt{2}$
• C. $2\log 2$
• D. $3\log \sqrt{2}$

Answer 1

The correct option is B $-\log \sqrt{2}$

Consider, $I={\int }_0^{\frac{\pi }{4}}\sqrt{\frac{1-sin2x}{1+sin2x}}\cdot dx$

$I={\int }_0^{\frac{\pi }{4}}\sqrt{\frac{(sinx-cosx{)}^2}{(sinx+cosx{)}^2}}\cdot dx$

$I={\int }_0^{\frac{\pi }{4}}\frac{(sinx-cosx)}{sinx+cosx}\cdot dx$

$I=-{\int }_0^{\frac{\pi }{4}}\frac{cosx-sinx}{sinx+cosx}dx$

$I=-{[log|sinx+cosx|]}_0^{\frac{\pi }{4}}$

$I=-\left[log\right(sin\frac{\pi }{4}+cos\frac{\pi }{4})-\log (\sin 0+\cos 0\left)\right)]$

$I=-[log(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})-\log 1]$

$I=-log(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})=-log\sqrt{2}$

So, ${\int }_0^{\frac{\pi }{4}}\sqrt{\frac{1-sin2x}{1+sin2x}}x=-log\left(\sqrt{2}\right)$