definite integral of a trigonometric function with log $$I=\int_0^{\frac\pi 2} \log(1+\tan x)\,dx$$

I attempted it by substituting $(\frac\pi 2-x)$ as $x$ but we get $\log(1+\cot x)$ not sure how to proceed... and on adding $$2I=\log[(1+\tan x)(1+\cot x)]$$ 


Another solution using Clausen function
Start with $$ I=\int \limits^{\frac{\pi }{2} }_{0}\ln\left( 1+\tan \left( x\right) \right) dx=\overbrace{\int \limits^{\frac{\pi }{2} }_{0}\ln\left( \sin \left( x\right) +\cos \left( x\right) \right) dx} \limits^{I_{1}}-\overbrace{\int \limits^{\frac{\pi }{2} }_{0}\ln\left( \cos \left( x\right) \right) } \limits^{I_{2}}$$
$$ I_{1}=\int \limits^{\frac{\pi }{2} }_{0}\ln\left( \sin \left( x\right) +\cos \left( x\right) \right) dx=\int \limits^{\frac{\pi }{2} }_{0}\ln\left( \sqrt{2} \left( \sin \left( \frac{\pi }{4} \right) \cos \left( x\right) +\cos \left( \frac{\pi }{4} \right) \sin \left( x\right) \right) \right) dx= \frac{-\pi }{4} \ln\left( 2\right) +\int \limits^{\frac{\pi }{2} }_{0}\ln\left( 2\sin \left( x+\frac{\pi }{4} \right) \right) dx=\frac{-\pi }{4} \ln\left( 2\right) +\frac{1}{2} \int \limits^{\frac{3\pi }{2} }_{\frac{\pi }{2} }\ln\left( 2\sin \left( \frac{t}{2} \right) \right) dt=\frac{-\pi }{4} \ln\left( 2\right) -\frac{1}{2} \text{Cl}_{2}\left( \frac{3\pi }{2} \right) +\frac{1}{2} \text{Cl}_{2}\left( \frac{\pi }{2} \right) =\boxed {\frac{-\pi }{4} \text{ln}\left( 2\right) +G}$$ 
$$   I_{2}=\int \limits^{\frac{\pi }{2} }_{0}\ln\left( \cos \left( x\right) \right) dx=\int \limits^{\frac{\pi }{2} }_{0}\text{ln}\left( \sin \left( x\right) \right) dx $$ 
$$ 2I_{2}=\int \limits^{\frac{\pi }{2} }_{0}\text{ln}\left( \sin \left( 2x\right) \right) dx-\int \limits^{\frac{\pi }{2} }_{0}\text{ln}\left( 2\right) dx=\frac{1}{2} \int \limits^{\pi }_{0}\text{ln}\left( \sin \left( u\right) \right) du-\frac{\pi }{2} \text{ln}\left( 2\right) =\overbrace{\int \limits^{\frac{\pi }{2} }_{0}\text{ln}\left( \sin \left( u\right) \right) du} \limits^{I_{2}}-\frac{\pi }{2} \text{ln}\left( 2\right) $$
$$ I_{2}=\boxed {\frac{-\pi }{2} \text{ln}\left( 2\right)}$$
We have  $I=I_{1}-I_{2} $ to get the final result 
$$ \color{purple}{\boxed {I=  G+\frac{\pi }{4} \text{ln}\left( 2\right) } }$$
