<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] Rewrite
$$
\begin{align}
\int_0^{\Large\frac{\pi}{2}}\tan x\ln(\sin x)\,dx=\int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sin x)\ dx.
\end{align}
$$
Let $\,u=\cos x$, then $\,du=-\sin x\,dx$. For $\,0 < x < \frac{\pi}{2}$, we have $\,0 < u < 1$. Now, the integral turns out to be
$$
\begin{align}
\int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sin x)\,dx&=\int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sqrt{1-\cos^2 x})\,dx\\
&=-\frac{1}{2}\int_0^1\frac{\ln(1-u^2)}{u}\,du.\tag1\\
\end{align}
$$
Next, use Maclaurin series for natural logarithm:
$$
\ln(1-u^2)=-\sum_{n=1}^\infty \frac{u^{2n}}{n}.\tag2\\
$$
Substitute $\,(2)$ to $\,(1)$, yield
$$
\begin{align}
\frac{1}{2}\int_0^1\frac{\ln(1-u^2)}{u}\,du&=-\frac{1}{2}\int_0^1\sum_{n=1}^\infty \frac{u^{2n}}{un}\,du\\
&=-\frac{1}{2}\sum_{n=1}^\infty\int_0^1 \frac{u^{2n-1}}{n}\,du\\
&=-\frac{1}{4}\sum_{n=1}^\infty \left.\frac{u^{2n}}{n^2}\right|_{u=0}^1\\
&=-\frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2}.\tag3
\end{align}
$$
The infinite series in $(3)$ is defined as Riemann zeta function $\,\zeta (2)=\dfrac{\pi^2}{6}$. Thus,
$$
\begin{align}
\int_0^{\Large\frac{\pi}{2}}\tan x\ln(\sin x)\,dx&=-\frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2}\\
&=-\frac{1}{4}\cdot \frac{\pi^2}{6}\\
&=\large\color{blue}{-\frac{\pi^2}{24}}.\\
\end{align}
$$
</s>