<s>[INST] Calculus Question: $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$ Can anyone help me to find  $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$? Any help would be appreciated. Thanks in advance.
 [/INST] Let
$$ I(a,b)=\int_0^{\pi/2}(\sin x)^{1-a}(\cos x)^{1-b}dx.$$
Then 
$$ I(a,b)=\frac{\Gamma \left(1-\frac{a}{2}\right) \Gamma \left(1-\frac{b}{2}\right)}{2
   \Gamma \left(-\frac{a}{2}-\frac{b}{2}+2\right)} $$
and hece
\begin{eqnarray}
\frac{\partial I(a,b)}{\partial a}&=&\frac{\Gamma \left(1-\frac{a}{2}\right) \Gamma \left(1-\frac{b}{2}\right)
   \left(\psi ^{(0)}\left(-\frac{a}{2}-\frac{b}{2}+2\right)-\psi
   ^{(0)}\left(1-\frac{a}{2}\right)\right)}{4 \Gamma
   \left(-\frac{a}{2}-\frac{b}{2}+2\right)}.
\end{eqnarray}
Thus
\begin{eqnarray}
\frac{\partial I(0,b)}{\partial a}&=&\frac{\Gamma \left(1-\frac{b}{2}\right) \left(\psi
   ^{(0)}\left(2-\frac{b}{2}\right)+\gamma \right)}{4 \Gamma
   \left(2-\frac{b}{2}\right)}.
\end{eqnarray}
Noting
$$ \Gamma(x)\approx\frac{1}{x} \text{ near }x=0 $$
we have
\begin{eqnarray}
\lim_{b\to2}\frac{\partial I(0,b)}{\partial a}&=&\lim_{b\to2}\frac{\Gamma \left(1-\frac{b}{2}\right) \left(\psi
   ^{(0)}\left(2-\frac{b}{2}\right)+\gamma \right)}{4 \Gamma
   \left(2-\frac{b}{2}\right)}\\
&=&\lim_{b\to2}\frac14 \frac{\psi^{(0)}(2-\frac{b}{2})+\gamma}{1-\frac{b}{2}}\\
&=&\lim_{b\to2}\frac14 \psi^{(1)}(2-\frac{b}{2})\\
&=&\frac{\pi^2}{24}.
\end{eqnarray}
So 
$$ \int_0^{\pi/2}\tan x\ln(\sin x)dx=-\frac{\pi^2}{24}. $$
</s>