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.
 A: 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,
$$
\begin{align}
\int_0^{\Large\frac{\pi}{2}}\tan x\ln(\sin x)\,dx&=-\frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2}\\
&=-\frac{1}{4}\cdot \frac{\pi^2}{6}\\
&=\large\color{blue}{-\frac{\pi^2}{24}}.\\
\end{align}
$$
A: Here is one line proof using beta function
$$\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx=\lim_{a\rightarrow 1,b\rightarrow 1^+}\frac{1}{2}\frac{\partial B}{\partial a}\left(\frac{1}{2}(a+1),\frac{1}{2}(b+1)\right)=-\frac{\pi^2}{24}$$
Q.E.D.
A: $$\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)log(\sin(x))}{\cos(x)}dx$$
Let $t=\cos(x), \;\ \sin(x)=\sqrt{1-t^{2}}, \;\ dx=\frac{-1}{\sqrt{1-t^{2}}}dt$
$$1/2\int_{0}^{1}\frac{log(1-t^{2})}{t}dt$$
This integral can be done using the series for $$log(1-t^{2})=-\sum_{k=1}^{\infty}\frac{t^{2k}}{k}$$, and is rather famous.
But, it evaluates to $$\frac{-\pi^{2}}{24}$$
A: Subst. $t=\sin{x}$, then the given integral becomes
\begin{align*}
  \int_0^1 \, \frac{t\, \log{t}}{1-t^2} dt &= \int_{0}^{1} \, \log{t} \, \sum_{k\ge 0} t^{2k+1} \, dt\\
&= \sum_{k\ge 0} \int_{0}^{1} \, \left(\log{t}\right)\, t^{2k+1}\, dt\\
&=  \sum_{k\ge 0} -\frac{1}{4 \, {\left(k^{2} + 2 \, k + 1\right)}}\\
&= -\frac{1}{4} \zeta{(2)}\\
&= -\frac{\pi^2}{24}
\end{align*}
In general, we can have
\begin{align*}
  \int_{0}^{\pi/2} \, \left(\log{\sin{x}}\right)^n\, \tan{x}\, dx = (-1)^n\, \frac{ n!\,  \zeta(n + 1)}{2^{n + 1}}
\end{align*}
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\overbrace{\color{#00f}{\large\int_{0}^{\pi/2}\tan\pars{x}\ln\pars{\sin\pars{x}}\,\dd x}}
^{\ds{t\ \equiv \sin\pars{x}}}\
=\ \overbrace{\int_{0}^{1}{t\ln\pars{t} \over 1 - t^{2}}\,\dd t}
^{\ds{t\ \equiv \expo{-\xi}}}\ =\
-\int_{0}^{\infty}{\xi\expo{-2\xi} \over 1 - \expo{-2\xi}}\,\dd\xi
\\[3mm]&=
-\,{1 \over 4}\int_{0}^{\infty}{\xi\expo{-\xi} \over 1 - \expo{-\xi}}\,\dd\xi
=\,{1 \over 4}\int_{0}^{\infty}\ln\pars{1 - \expo{-\xi}}\,\dd\xi
={1 \over 4}\int_{0}^{\infty}
\pars{-\sum_{n = 1}^{\infty}{\expo{-n\xi} \over n}}\,\dd\xi
\\[3mm]&=-\,{1 \over 4}\sum_{n = 1}^{\infty}{1 \over n}\
\underbrace{\int_{0}^{\infty}\expo{-n\xi}\,\dd\xi}_{\ds{=\ {1 \over n}}}
=-\,{1 \over 4}\ \underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}}
_{\ds{=\ {\pi^{2} \over 6}}} = \color{#00f}{\Large -\,{\pi^{2} \over 24}}
\end{align}
A: I just finished this answer and noticed that equation $(6)$ there answers this question. I did a small amount of customization for this question.
Let $t=\sin(x)$ and $e^{-u/2}=t$
$$
\begin{align}
\int_0^{\pi/2}\tan(x)\log(\sin(x))\,\mathrm{d}x
&=\int_0^1\frac{t}{1-t^2}\log(t)\,\mathrm{d}t\\
&=-\frac14\int_0^\infty\frac{e^{-u}}{1-e^{-u}}u\,\mathrm{d}u\\
&=-\frac14\int_0^\infty\sum_{k=1}^\infty ue^{-ku}\,\mathrm{d}u\\
&=-\frac14\sum_{k=1}^\infty\frac1{k^2}\\
&=-\frac{\pi^2}{24}
\end{align}
$$
This is only slightly different from Felix Marin's answer, using the series for $\dfrac{x}{1-x}$ instead of $\log(1-x)$.
A: 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 
$$ \int_0^{\pi/2}\tan x\ln(\sin x)dx=-\frac{\pi^2}{24}. $$
