evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$ 
Compute the indefinite integral
  $$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$

My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\frac{\pi}{4}+x\right)
$$
This changes the integral to
$$
\int \cos (2x)\cdot \ln \left(\tan \left(\frac{\pi}{4}+x\right)\right)\,dx
$$
Now let $t=\left(\frac{\pi}{4}+x\right)$ such that $dx = dt$. Then the integral with changed variables becomes
$$
\begin{align}
\int \cos \left(2t-\frac{\pi}{2}\right)\cdot \ln (\tan t)dt &= \int \sin (2t)\cdot \ln (\tan t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \frac{\sec^2(t)}{\tan t}\cdot \cos (2t)\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \cot (2t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\ln \left|\sin (2t)\right|
\end{align}
$$
where $t=\displaystyle \left(\frac{\pi}{4}+x\right)$.
Is this solution correct? Is there another method for finding the solution?

Integrate by parts: $\int udv=uv-\int v du$, where
$$u=\ln\frac{\cos x+\sin x}{\cos x-\sin x}\Rightarrow du=\frac{\frac{(\cos x-\sin x)(-\sin x+\cos x)-(\cos x+\sin x)(-\sin x +\cos x)  }{(\cos x-\sin x)^2}}{\frac{\cos x+\sin x}{\cos x-\sin x}}=...=\frac{2}{\cos 2x}dx$$
and
$$ dv=\cos 2x dx \Rightarrow v=\frac{1}{2}\sin 2x.$$
Then,
$$\int \cos 2x \ln(\frac{\cos x+\sin x}{\cos x-\sin x}) dx=\frac{1}{2}\sin 2x \ln\frac{\cos x+\sin x}{\cos x-\sin x}-\int \frac{1}{2}\sin 2x \frac{2}{\cos 2x} dx=$$
$$=\frac{1}{2}\sin 2x\ln \frac{\cos x+\sin x}{\cos x-\sin x}-\int \tan 2x dx= $$
$$=\frac{1}{2}\sin 2x \cdot\ln\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)-\frac{1}{2}\ln|\sec 2x|+c. $$
