Question

$\int \left(\frac{1-\cos 2x}{1+\cos 2x}\right)dx$

Answer 1

Given that,

$\begin{array}{c}\int \frac{1-\cos \left(2x\right)}{1+\cos \left(2x\right)}dx\\ \int \frac{1-\cos \left(2x\right)}{1+\cos \left(2x\right)}=\frac{2}{1+\cos \left(2x\right)}-1\\ \int \frac{2}{1+\cos \left(2x\right)}-1dx\end{array}$

We know that,

$\begin{array}{c}\int f\left(x\right)\pm g\left(x\right)=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\ \int \frac{2}{1+\cos \left(2x\right)}-\int 1dx\\ \int \frac{2}{1+\cos \left(2x\right)}=\tan \left(x\right)\\ \int 1dx=x\\ =\tan \left(x\right)-x\end{array}$

Then,

We get $=\tan \left(x\right)-x+C$