Computing $\int_{0}^{\pi/2}\cos(x)\ln(\tan(x))dx$ 
Compute $\int_{0}^{\pi/2}\cos(x)\ln(\tan(x))dx$

It is easy to check this improper integral converges. One also notes that $\int_{0}^{\pi/2}\cos(x)\ln(\tan(x))dx=-\int_{0}^{\pi/2}\sin(x)\ln(\tan(x))dx$.
It is possible to find an antiderivative for $\cos(x)\ln(\tan(x))$ using integration by parts, but I'm looking for a nicer way, that would involve only substitution or other tricks.

For $x \in [0,\frac{\pi}{2}]$, we have that $\cos x = \sqrt{1 - \sin^2 x}$, so $$ \log (\tan x) = \log (\sin x) - \log (\cos x) = \log (\sin x) - \frac{1}{2} \log ( 1 - \sin^2 x)$$ Then make the substitution $u = \sin x$. Since $du = \cos x \;dx$, we get that $$\int_{0}^{\pi/2}\cos(x)\log(\tan(x))dx$$ $$\int_{0}^{\pi/2}\cos(x)\log(\sin(x))dx-\frac{1}{2} \int_{0}^{\pi/2}\cos(x)\log(1-\sin^2 x)dx$$ $$\int_0^1 \log u\; du -\frac{1}{2} \int_0^1 \log(1-u^2)\; du $$$$= \int_0^1 \log u \;du -\frac{1}{2} \int_0^1 \log(1-u)+\log(1+u)\; du$$ and this can be easily integrated, since the antiderivative of $\log y$ is equal to $y \log y -y +C$.
