Question

The value of $\int \sqrt{\frac{1-\sin 2x}{1-\cos 2x}}dx$ is

• A. $\frac{1}{\sqrt{2}}(x-\cos x+c)$
• B. $\frac{1}{\sqrt{2}}(\log |\sin x|+c)$
• C. $\frac{1}{\sqrt{2}}(x+\tan x\sec x+c)$
• D. $\frac{1}{\sqrt{2}}(x-\log |\sin x|+c)$

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}(x-\log |\sin x|+c)$

$\int \sqrt{\frac{1-\sin 2x}{1-\cos 2x}}dx=\int \sqrt{\frac{({\sin }^{2}x+{\cos }^{2}x)-\left(2\sin x\cos x\right)}{({\sin }^{2}x+{\cos }^{2}x)-({\cos }^{2}x-{\sin }^{2}x)}}dx=\int \sqrt{\frac{(\sin x-\cos x{)}^2}{2{\sin }^{2}x}}dx=\frac{1}{\sqrt{2}}\int \frac{\sin x-\cos x}{\sin x}dx=\frac{1}{\sqrt{2}}\int 1-\cot xdx=\frac{1}{\sqrt{2}}(x-\log |\sin x|+c)$