\(\int\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}dx=\)
Answer & explanation
Correct answer: option 2
\(\int\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}dx=\int\sqrt{\frac{\sin^2x}{\cos^2}}dx\)
$=\int\tan x\,dx$
$=-\log|\cos x|+c$
\(\int\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}dx=\)
Correct answer: option 2
\(\int\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}dx=\int\sqrt{\frac{\sin^2x}{\cos^2}}dx\)
$=\int\tan x\,dx$
$=-\log|\cos x|+c$