How to evaluate $I=\int_0^{\frac{\pi}{2}}\sin^2x\ln(\sin^2(\tan x))dx$ $$I=\int_0^{\frac{\pi}{2}}\sin^2x\ln(\sin^2(\tan x))dx \hspace{15mm}(1)$$
Now, using definite integral property of $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$ $$I=\int_0^{\frac{\pi}{2}}\cos^2x\ln(\sin^2(\cot x))dx\hspace{15mm}(2)$$
After $\tan x=t$ substitution in $(1)$ and $\cot x=m$ in $(2)$, to get: $$I=\int_0^{\frac{\pi}{2}}\frac{t^2}{(1+t^2)^2}\ln(\sin^2 t)dt=\int_0^{\frac{\pi}{2}}\frac{m^2}{(1+m^2)^2}\ln(\sin^2 m)dm$$
After variable change and addition, I get:
$$2I=\int_0^{\frac{\pi}{2}}\frac{x^2}{(1+x^2)^2}\ln(\sin^4 x)dx\implies \frac{I}{2}=\int_0^{\frac{\pi}{2}}\frac{x^2}{(1+x^2)^2}\ln(\sin x)dx$$
How could I proceed? Any other solutions which happen to be more efficient/simple?
 A: $$I=\int_0^{\frac{\pi}{2}}\cos^2x\ln(\sin^2(\tan x))dx$$
Use the substitution $t = \tan(x)$
$$I=2\int_0^{\infty}\frac{\ln|\sin t|}{(1+t^2)^2}dt $$
Then use the fourier expansion of $\log|\sin x|$
$$I=-2\sum\frac{1}{k}\int_0^{\infty}\frac{\cos(2kt)}{(1+t^2)^2}dt-2\int_0^{\infty}\frac{\log(2)}{(1+t^2)^2}dt  $$
For the second integral 
$$2\int_0^{\infty}\frac{\log(2)}{(1+t^2)^2}$$ 
We can use the beta function to deduce that 
$$2\int_0^{\infty}\frac{\log(2)}{(1+t^2)^2} =\frac{\pi}{2}\log(2) $$ 
For the second integral 
$$\int_0^{\infty}\frac{\cos(2kt)}{(1+t^2)^2}dt = Re\int_0^{\infty}\frac{e^{2kit}}{(1+t^2)^2}dt$$
It can be showing by complex analysis that 
$$\int_0^{\infty}\frac{\cos(2kt)}{(1+t^2)^2}dt  = \frac{\pi}{4}e^{-2k}(1+2k) $$
Finally we have the integral 
$$I=-\frac{\pi}{2}\sum \frac{e^{-2k}(1+2k)}{k }  -\frac{\pi}{2}\log(2)$$
The sum can be computed using 
$$-\log(1-x) = \sum \frac{x^k}{k}$$
Hence we deduce that 

$$I=-\frac{\pi}{2}\left\{\log\left(\frac{2}{e^2-1} \right) +\frac{2e^ 2}{e^2-1}\right\}$$


ADDENDUM 
As required A proof using complex analysis for
$$\int_0^{\infty}\frac{\cos(2kt)}{(1+t^2)^2}dt$$
This can be done by considering 
$$F(z) = \frac{e^{2ki z}}{(1+z^2)^2}$$
The poles are $z=\pm i$, then consider half a circle in the upper half plane. 

Note there is only one pole of order 2 inside the contour. Hence by the residue theorem 
$$\int_C \frac{e^{2ki z}}{(1+z^2)^2}+\int_{-\infty}^0\frac{e^{2ki x}}{(1+x^2)^2} \,dx+ \int^{\infty}_0\frac{e^{2ki x}}{(1+x^2)^2}\,dx = 2\pi i \text{Res} \frac{e^{2ki z}}{(1+z^2)^2}|_{z = i} $$
The integral along the arc converges to 0 
$$2\int^{\infty}_0\frac{\cos(2k x)}{(1+x^2)^2} \,dx = 2\pi i \text{Res} \frac{e^{2ki z}}{(1+z^2)^2}|_{z =  i} $$
The pole of order 2 can be calculated using 
$$\lim_{z \to i} \frac{d}{dz} (z-i)^2 \frac{e^{2ki z}}{(1+z^2)^2}$$
This simplifies to 
$$2\int^{\infty}_0\frac{\cos(2k x)}{(1+x^2)^2} \,dx = \frac{\pi}{2}e^{-2k}(1+2k)$$
$$\int^{\infty}_0\frac{\cos(2k x)}{(1+x^2)^2} \,dx = \frac{\pi}{4}e^{-2k}(1+2k)$$
For many other approaches see Calculating the integral $\int_{0}^{\infty} \frac{\cos x}{1+x^2}\mathrm{d}x$ without using complex analysis
