Evaluating $\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx$ What starting point would you recommend me for the one below?
$$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx $$
EDIT
Thanks to Felix Marin, we know the integral evaluates to  
$$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$
 A: Let $I = \int_0^1 x \tan(\pi x) \log(\sin(\pi x)) dx$. We have
$$I= \int_0^1 \dfrac{x \tan(\pi x)}2 \log(1-\cos^2(\pi x))dx$$
Hence,
$$2I = -\sum_{k=1}^{\infty}\dfrac1k\int_0^1 x \tan(\pi x) \cos^{2k}(\pi x)dx \,\,\,\,\,\,\,\, \spadesuit$$
\begin{align}
\int_0^1 x \tan(\pi x) \cos^{2k}(\pi x)dx & = \int_0^1 x\sin(\pi x) \cos^{2k-1}(\pi x)dx\\
& = \int_0^{1/2} x\sin(\pi x) \cos^{2k-1}(\pi x)dx + \int_{1/2}^1 x\sin(\pi x) \cos^{2k-1}(\pi x)dx\\
& = \int_0^{1/2} x\sin(\pi x) \cos^{2k-1}(\pi x)dx - \int_0^{1/2} (1-x)\sin(\pi x) \cos^{2k-1}(\pi x)dx\\
& = 2 \int_0^{1/2} x\sin(\pi x) \cos^{2k-1}(\pi x)dx - \int_0^{1/2}\sin(\pi x) \cos^{2k-1}(\pi x)dx\\
& = \frac2{\pi^2}\int_0^1 t^{2k-1} \arccos(t) dt - \frac1{\pi}\int_0^1 t^{2k-1} dt\\
& = \frac1{2\pi^{3/2}} \dfrac{\Gamma(k+1/2)}{k^2\Gamma(k)} - \frac1{2 \pi k}
\end{align}
Use this in $\spadesuit$ and the approrpriate Taylor series will give you the answer. I am away now since I have a class to teach. Will add more details if needed sometime later.

Updated
We have
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{\Gamma(k)} x^{k-1} = \dfrac{\sqrt{\pi}}2 \dfrac1{(1-x)^{3/2}}$$
Integrate the above once and divide by $x$ to obtain
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k\Gamma(k)} x^{k-1} = \dfrac{\sqrt{\pi}}{x(1-x)^{1/2}} + \frac{c_1}x$$
To obtain $c_1$, take $\lim_{x \to 0}$, the left hand side is $\Gamma(1/2)$. For the right hand side limit to even exist, we need $c_1 = -\sqrt{\pi}$. Hence,
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k\Gamma(k)} x^{k-1} = \dfrac{\sqrt{\pi}}{x(1-x)^{1/2}} - \frac{\sqrt{\pi}}x$$
Integrate again and divide by $x$ to obtain
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k^2\Gamma(k)} x^{k-1} = \sqrt{\pi} \left(\dfrac{\log\left(1-\sqrt{1-x} \right) - \log\left(1+\sqrt{1-x} \right)}x \right) - \sqrt{\pi}\frac{\log(x)}x + \frac{c_2}x$$
To obtain $c_2$, take limit $x \to 0$. For the right hand side limit to even exist, we need $c_2 = 2\sqrt{\pi} \log(2)$. Hence,
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k^2\Gamma(k)} x^{k-1} = \dfrac{2\sqrt{\pi}\log2-2\sqrt{\pi} \log\left(1+\sqrt{1-x}\right)}{x}$$
Integrate again to obtain
\begin{align}
\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k^3\Gamma(k)} x^{k} = \frac{\sqrt{\pi}}{2} \left(2\text{Li}_2(1/2(\sqrt{1-x}-1)) - 2\text{Li}_2(1/2(\sqrt{1-x}+1))\\
 + \left(\log(1-\sqrt{1-x}) - \log(1+\sqrt{1-x}) \right)\left(\log(1-\sqrt{1-x}) + \log(1+\sqrt{1-x}) -2\log2\right)\right) - \sqrt{\pi} \frac{\log^2(x)}2 + 2\sqrt{\pi} \log2 \log(x) + c_3
\end{align}
Now plug in $x=1$ in the above to get the sum of the first series and the sum of the second series is $$\sum_{k=1}^{\infty} \dfrac1{k^2} = \dfrac{\pi^2}6$$
I shall add more details if needed.
A: \begin{align}
\mathcal{I}=&\int^1_0x\tan(\pi x)\ln(\sin(\pi x))\ {\rm d}x\\
=&\left(\int^{1/2}_0+\int_{1/2}^1\right)x\tan(\pi x)\ln(\sin(\pi x))\ {\rm d}x\tag1\\
=&\int^{1/2}_0(2x-1)\tan(\pi x)\ln(\sin(\pi x))\ {\rm d}x\tag2\\
=&-\frac{2}{\pi^2}\int^{\pi/2}_0x\cot{x}\ln(\cos{x})\ {\rm d}x\tag3\\
=&-\frac{1}{\pi}\int^{\pi/2}\tan{x}\ln(\sin{x})\ {\rm d}x+\frac{2}{\pi^2}\int^{\pi/2}_0x\tan{x}\ln(\sin{x})\ {\rm d}x\tag4\\
=&\frac{2}{\pi^2}\int^{\pi/2}_0\ln(\sin{x})\ln(\cos{x})\ {\rm d}x-\frac{2}{\pi^2}\int^{\pi/2}_0x\tan{x}\ln(\sin{x})\ {\rm d}x\tag5\\
=&\frac{1}{\pi^2}\int^{\pi/2}_0\ln(\sin{x})\ln(\cos{x})\ {\rm d}x-\frac{1}{2\pi}\int^{\pi/2}_0\tan{x}\ln(\sin{x})\ {\rm d}x\tag6\\
=&\frac{1}{8\pi^2}\frac{\partial^2{\rm B}}{\partial a\partial b}\left(\frac{1}{2},\frac{1}{2}\right)-\frac{1}{8\pi}\frac{\partial{\rm B}}{\partial b}\left(0^+,1\right)\tag7\\
=&\frac{1}{8\pi^2}\left(4\pi\ln^2{2}-\frac{\pi^3}{6}\right)-\frac{1}{8\pi}\left(-\frac{\pi^2}{6}\right)=\boxed{\Large{\color{red}{\dfrac{\ln^2{2}}{2\pi}}}}\\
\end{align}

Explanation:
$(1)$: Split the integral at $\displaystyle\frac12$.
$(2)$: Substituted $\displaystyle x\mapsto1-x$ in the second integral.
$(3)$: Substituted $\displaystyle x\mapsto\frac{1}{2}-\frac{x}{\pi}$. 
$(4)$: Substituted $\displaystyle x\mapsto\frac{\pi}{2}-x$. 
$(5)$: Integrated by parts. 
$(6)$: Took the average of $(4)$ and $(5)$. 
$(7)$: $\displaystyle {\rm B}(a,b)=2\int^{\pi/2}_0\sin^{2a-1}{x}\cos^{2b-1}{x}\ {\rm d}x$
A: 
We can get the serials Ana calculated.
A: Integrating by parts, we get
$$
\begin{align}
&\int_0^1x\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x\\
&=\int_0^{1/2}(2x-1)\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x\tag{1a}
\\
&=-\frac1\pi\int_0^{1/2}(2x-1)\log(\sin(\pi x))\,\mathrm{d}\log(\cos(\pi x))\tag{1b}\\
&=\small\frac2\pi\int_0^{1/2}\log(\sin(\pi x))\log(\cos(\pi x))\,\mathrm{d}x
+\int_0^{1/2}(2x-1)\cot(\pi x)\log(\cos(\pi x))\,\mathrm{d}x\tag{1c}\\
&=\small\frac2\pi\int_0^{1/2}\log(\sin(\pi x))\log(\cos(\pi x))\,\mathrm{d}x
+\int_0^{1/2}(-2x)\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x\tag{1d}\\
&=\small\frac1\pi\int_0^{1/2}\log(\sin(\pi x))\log(\cos(\pi x))\,\mathrm{d}x
-\frac12\int_0^{1/2}\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x\tag{1e}\\
\end{align}
$$
Explanation:
$\text{(1a)}$: Subtract $\frac12\int_0^1\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x=0$ and use the symmetry of $x\mapsto1-x$
$\text{(1b)}$: Apply $\mathrm{d}\log(\cos(\pi x))=\pi\tan(\pi x)\,\mathrm{d}x$
$\text{(1c)}$: Integrate by Parts
$\text{(1d)}$: Substitute $x\mapsto\frac12-x$
$\text{(1e)}$: Average $\text{(1a)}$ and $\text{(1d)}$
Using
$$
\log(\sin(x))=-\log(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k}\tag{2}
$$
$$
\log(\cos(x))=-\log(2)-\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}\tag{3}
$$
we get
$$
\begin{align}
&\int_0^{1/2}\log(\sin(\pi x))\log(\cos(\pi x))\,\mathrm{d}x\\
&=\color{#C00000}{\int_0^{1/2}(-\log(2))^2\,\mathrm{d}x}\\
&-\color{#00A000}{\log(2)\int_0^{1/2}\sum_{k=1}^\infty\left[-1-(-1)^k\right]\frac{\cos(2\pi kx)}{k}\,\mathrm{d}x}\\
&+\color{#0000FF}{\int_0^{1/2}\sum_{j=1}^\infty\sum_{k=1}^\infty(-1)^k\frac{\cos(2\pi jx)}{j}\frac{\cos(2\pi kx)}{k}\,\mathrm{d}x}\\
&=\color{#C00000}{\frac{\log(2)^2}2}-\color{#00AA00}{0}+\color{#0000FF}{\frac14\sum_{k=1}^\infty\frac{(-1)^k}{k^2}}\\
&=\frac{\log(2)^2}2-\frac{\pi^2}{48}\tag{4}
\end{align}
$$
since $\int_0^{1/2}\cos(2\pi jx)\cos(2\pi kx)\,\mathrm{d}x=0$ when $j\ne k$.
Substituting $t=\sin(\pi x)$ and $e^{-u/2}=t$, we get
$$
\begin{align}
\int_0^{1/2}\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x
&=\frac1\pi\int_0^1\frac{t}{1-t^2}\log(t)\,\mathrm{d}t\\
&=-\frac1{4\pi}\int_0^\infty\frac{e^{-u}}{1-e^{-u}}u\,\mathrm{d}u\\
&=-\frac1{4\pi}\int_0^\infty\sum_{k=1}^\infty ue^{-ku}\,\mathrm{d}u\\
&=-\frac1{4\pi}\sum_{k=1}^\infty\frac1{k^2}\\
&=-\frac{\pi}{24}\tag{6}
\end{align}
$$
Combining $(1)$, $(4)$, and $(6)$ yields
$$
\int_0^1x\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x
=\frac{\log(2)^2}{2\pi}\tag{7}
$$
A: $\def\Li{{\rm{Li}_2\,}}$Denote the considered integral as $I$ and set $y=\pi x$, we have
\begin{equation}
I=\frac{1}{\pi^2}\int_0^\pi \frac{y\sin y}{\cos y}\,\ln(\sin y)\,dy
\end{equation}
Perform integration by parts by taking $u=y$, we have
\begin{align}
I&=-\left.\frac{y}{2\pi^2}\int\frac{\ln\left(1-\cos^2y\right)}{\cos y}\,d(\cos y)\right|_0^\pi+\frac{1}{2\pi^2}\int_0^\pi\int\frac{\ln\left(1-\cos^2y\right)}{\cos y}\,d(\cos y)\,dy\\
&=\left.\frac{y}{4\pi^2}\Li\left(\cos^2y\right)\,\right|_0^\pi-\frac{1}{4\pi^2}\int_0^\pi\Li\left(\cos^2y\right)\,dy\\
&=\frac{\Li\left(1\right)}{4\pi}-\frac{1}{4\pi^2}\int_0^{\pi}\sum_{k=1}^\infty\frac{\cos^{2k}y}{k^2}\,dy\quad\Rightarrow\quad\color{red}{\mbox{use series representation of dilogarithm}}\\
&=\frac{\pi}{24}-\frac{1}{4\pi^2}\sum_{k=1}^\infty\int_0^{\pi}\frac{\cos^{2k}y}{k^2}\,dy\quad\Rightarrow\quad\color{red}{\mbox{justified by Fubini-Tonelli theorem}}\\
&=\frac{\pi}{24}-\frac{1}{2\pi^2}\sum_{k=1}^\infty\frac{1}{k^2}\int_0^{\pi/2}\cos^{2k}y\,\,dy\quad\Rightarrow\quad\color{red}{\mbox{by symmetry argument}}\\
&=\frac{\pi}{24}-\frac{1}{4\pi^2}\sum_{k=1}^\infty\frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{k^2\,\Gamma\left(k+1\right)}\quad\Rightarrow\quad\color{red}{\mbox{Wallis' integrals}}\\
&=\frac{\pi}{24}-\frac{1}{4\pi}\sum_{k=1}^\infty\frac{(2k)!}{4^k\,k^2\,(k!)^2}\tag1
\end{align}
$\def\arctanh{{\rm{\,arctanh}\,}}$Here is the tedious part (and also the hardest part). I use Mathematica to help me out to find generating function of the following series. Let us start with
\begin{equation}
\sum_{k=0}^\infty\frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)x^k}{\Gamma\left(k+1\right)}=\pi\sum_{k=0}^\infty\frac{(2k)!\,\,x^k}{4^k\,(k!)^2}=\frac{\pi}{\sqrt{1-x}}
\end{equation}
Divide by $x$ and then integrate it, we have
\begin{equation}
\sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k\,(k!)^2}=\int\left[\frac{1}{x\,\sqrt{1-x}}-\frac{1}{x}\right]\,dx=-2\arctanh\left(\sqrt{1-x}\,\right)-\ln x+C_1
\end{equation}
Taking the limit as $x\to0$, we obtain
\begin{equation}
C_1=\lim_{x\to0}\left(2\arctanh\left(\sqrt{1-x}\,\right)+\ln x\right)=\ln4
\end{equation}
Hence
\begin{equation}
\sum_{k=1}^\infty\frac{(2k)!}{4^k\,k\,(k!)^2}x^k=-2\arctanh\left(\sqrt{1-x}\,\right)-\ln x+\ln4
\end{equation}
Repeat the process once more, we obtain
\begin{align}
\sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k^2\,(k!)^2}=&\,-2\int\frac{\arctanh\left(\sqrt{1-x}\,\right)}{x}\,dx-\int\frac{\ln x}{x}\,dx+\ln4\int\frac{dx}{x}\\
=&\,\,2\arctanh\left(\sqrt{1-x}\,\right)\left[\arctanh\left(\sqrt{1-x}\,\right)-2\ln\left(\frac{\sqrt{1-x}+1}{2}\right)\right]\\&\,-2\,\Li\left(\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right)-\frac{\ln^2x}{2}+\ln4\ln x+C_2\\
\end{align}
Taking the limit as $x\to0$, we obtain
\begin{equation}
C_2=-2\ln^22
\end{equation}
Hence
\begin{align}
\sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k^2\,(k!)^2}
=&\,\,2\arctanh\left(\sqrt{1-x}\,\right)\left[\arctanh\left(\sqrt{1-x}\,\right)-2\ln\left(\frac{\sqrt{1-x}+1}{2}\right)\right]\\&\,-2\,\Li\left(\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right)-\frac{\ln^2x}{2}+\ln4\ln x-2\ln^22\tag2
\end{align}
Thus, by putting $x=1$ to $(2)$ then $(1)$ becomes
\begin{equation}
I=\frac{\pi}{24}-\frac{1}{4\pi}\left(\frac{\pi^2}{6}-2\ln^2 2\right)=\frac{\ln^22}{2\pi}
\end{equation}
and we are done.
