Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$ Please help me to solve this integral:
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx.$$
I managed to calculate an indefinite integral of the left part:
$$\int\frac{\cos x}{\sin x}x\ \mathrm dx=\ x\log(2\sin x)+\frac{1}{2} \Im\ \text{Li}_2(e^{2\ x\ i}),$$
where $\Im\ \text{Li}_2(z)$ denotes the imaginary part of the dilogarithm. The corresponding definite integral $$\int_0^\pi\frac{\cos x}{\sin x}x\ \mathrm dx$$ diverges. So, it looks like in the original integral summands compensate each other's singularities to avoid divergence.
I tried a numerical integration and it looks plausible that
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx\stackrel{?}{=}\pi \log 54,$$
but I have no idea how to prove it.
 A: Begin with $u$-substitution using 
$$\begin{align}
u & =x\\
dv & = \frac{3\cos(x)+\sqrt{8+\cos^2(x)}}{\sin(x)}\,dx\end{align}$$
so that $du=dx$, and my CAS tells me (which I suppose could be verified through differentiation and identities) that
$$\begin{align}
v & = \sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))
\end{align}$$
Now we have 
$$\begin{align}
\left[x\left(\sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))\right)\right]_0^\pi\\
-\int_0^\pi\left(\sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))\right)dx
\end{align}$$
and most of the integral part can be evaluated by taking advantage of symmetry about $\pi/2$:
$$\begin{align}
\left[x\left(\sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))\right)\right]_0^\pi\\
-3\int_0^\pi\ln(1-\cos(x))dx
\end{align}$$ 
($\sinh^{-1}$ is odd and $\cos(x)$ has odd symmetry about $\pi/2$. For the logarithmic term, the input to $\ln()$ at $x$ is the reciprocal of the input at $\pi/2-x$.) 
Some of the nonintegral-part can be cleanly evaluated:
$$\begin{align}
\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\left[3x\ln(1-\cos(x))\right]_0^\pi\\
-3\int_0^\pi\ln(1-\cos(x))dx
\end{align}$$ 
and now moving the "unclean" part back into an integral:
$$\begin{align}
\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\int_0^\pi\left(3\ln(1-\cos(x))+\frac{3x\sin(x)}{1-\cos(x)}\right)\,dx\\
-3\int_0^\pi\ln(1-\cos(x))dx\\
=\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\int_0^\pi\frac{3x\sin(x)}{1-\cos(x)}\,dx
\end{align}$$ 
My CAS says this is
$$\begin{align}
\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\pi\ln(64)
\end{align}$$
which is the only thing the CAS does that I don't quite get. But it's nothing special about endpoints: even WA can give an antiderivative if we can use the dilogarithm. It looks like an integral that might even appear somewhere on this site. A conversion of the arcsinh and logarithm rules yields
$$\begin{align}
\pi\ln(2^{-\frac{1}{2}})+\pi\ln\left(\frac{27}{8^{3/2}}\right)+\pi\ln(64)=\pi\ln(54)
\end{align}$$
A: Here's one way to go. 
First, note that 
$$\begin{eqnarray*}
\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx
&=& \int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx 
+\int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx.
\end{eqnarray*}$$
For now I'll simply claim that 
\begin{equation*}
\int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx = \pi\log 64.\tag{1}
\end{equation*}
(I would be surprised if this integral has not been handled somewhere on this site.)
But
$$\begin{eqnarray*}
\int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx
&=& \int_0^\pi\frac{3x}{\sin x} \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k}} \sin^{2k}x \ \mathrm dx \\
&=& \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k-1}}
\int_0^\pi x  \sin^{2k-1}x \ \mathrm dx \\
&=& \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k-1}}
\frac{\pi^{3/2}\Gamma(k)}{2\Gamma(k+1/2)} \\
&=& -\pi \sum_{k=1}^\infty \frac{1}{3^{2k-1}2k(2k-1)} \\
&=& -\pi \log \frac{32}{27}.
\end{eqnarray*}$$
(The last sum can be found by standard methods.
Schematically, $\sum \frac{a^{2k-1}}{2k(2k-1)} = \sum \int {\mathrm da} \frac{a^{2k-2}}{2k}$.) 
Thus, the integral is $\pi \log 54$ as claimed. 

Proof of (1): 
We have 
$$\begin{eqnarray*}
\int_0^\pi \frac{3x(1+\cos x)}{\sin x} \ \mathrm dx 
&=& \int_{0^+}^\pi \frac{3x(1+\cos x)}{\sin x} \ \mathrm dx \\
&=& 3\int_{0^+}^\pi x \cot\frac{x}{2} \ \mathrm dx
\hspace{5ex}\textrm{(double angle formulas)} \\
&=& 12 \int_{0^+}^{\pi/2} t\cot t \ \mathrm dt
\hspace{5ex} (t = x/2) \\
&=& -12\int_{0^+}^{\pi/2} \log\sin t \ \mathrm dt
\hspace{5ex}\textrm{(integrate by parts)} \\
&=& -6\int_{0^+}^{\pi/2} \log\sin^2 t \ \mathrm dt \\
&=& -6\int_{0^+}^{\pi/2} \log(1-\cos^2 t) \ \mathrm dt \\
&=& 6 \int_{0^+}^{\pi/2} \sum_{k=1}^\infty \frac{1}{k}\cos^{2k}t \ \mathrm dt
\hspace{5ex}\textrm{(series for log)} \\
&=& 6\sum_{k=1}^\infty \frac{1}{k} \int_{0^+}^{\pi/2} \cos^{2k}t \ \mathrm dt 
\hspace{5ex} \textrm{(Tonelli's theorem)}\\
&=& 6\sum_{k=1}^\infty \frac{1}{k}
    \frac{\sqrt{\pi}\Gamma(k+1/2)}{2\Gamma(k+1)}  \\
&=& 3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1}\frac{2k-1}{k} \\
&=& \pi \log 64.
\end{eqnarray*}$$
Note that 
$$\begin{eqnarray*}
6\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1} 
&=& -6\pi \left[\sum_{k=0}^\infty {1/2 \choose k}(-1)^{k} - 1\right] \\
&=& -6\pi[(1-1)^{1/2} - 1] \\
&=& 6\pi 
\end{eqnarray*}$$
and 
$$\begin{eqnarray*}
-3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1} \frac{1}{k}
&=& 3\pi \sum_{k=1}^\infty {1/2\choose k}(-1)^k  \int_0^1 x^{k-1} \ \mathrm dx \\
&=& 3\pi \int_0^1 \frac{1}{x}  \left[
\sum_{k=0}^\infty {1/2\choose k}(-1)^k x^{k} -1
\right] \ \mathrm dx \\
&=& 3\pi \int_0^1 \frac{1}{x}  \left(
\sqrt{1-x} -1
\right) \ \mathrm dx \\
&=& 3\pi(-2+\log 4) \\
&=& -6\pi + \pi\log 64.
\end{eqnarray*}$$
A: Let 
$$y=\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x},$$
then, solving this with respect to $x$, we get
$$x=\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}.$$
So,
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx=\int_0^\infty\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}\right)\mathrm dy.$$
The latter integral can be solved by Mathematica and yields $$\pi\log54.$$

Of course, we want to prove that the result returned by Mathematica is correct.

The following statement is provably true, that can be checked directly by taking derivatives of both sides:
$$\int\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}\right)\mathrm dy =\\ \frac{1}{2} i \left(2 \text{Li}_2\left(\frac{iy}{8}+\frac{1}{2}\right)+\text{Li}_2\left(\frac{iy}{6}+\frac{1}{3}\right)+2\text{Li}_2\left(\frac{iy}{6}+\frac{2}{3}\right)+\text{Li}_2\left(\frac{iy}{4}+\frac{1}{2}\right)+\text{Li}_2\left(\frac{2i}{y-2 i}\right)-\text{Li}_2\left(-\frac{2 i}{y+2i}\right)-\text{Li}_2\left(-\frac{1}{6} i (y+2i)\right)-\text{Li}_2\left(-\frac{1}{4} i (y+2i)\right)-2 \left(-\text{Li}_2\left(-\frac{2i}{y-4 i}\right)+\text{Li}_2\left(\frac{2 i}{y+4i}\right)+\text{Li}_2\left(-\frac{1}{8} i (y+4i)\right)+\text{Li}_2\left(-\frac{1}{6} i (y+4i)\right)\right)\right)+\pi  \left(\frac{1}{2}\log \left(3 \left(y^2+4\right)\right)+\log\left(\frac{3}{64}\left(y^2+16\right)\right)\right)+\log \left(4\left(y^2+4\right)\right) \arctan\left(\frac{y}{4}\right)-\left(\log576-2\log \left(y^2+16\right)\right) \arctan\left(\frac{4}{y}\right)+\log\left(y^2+4\right) \text{arccot}\left(\frac{6y}{8-y^2}\right)-\arctan\left(\frac{2}{y}\right)\log12 +\arctan\left(\frac{y}{2}\right)\log2$$
The remaining part is to calculate $\lim\limits_{y\to0}$ and $\lim\limits_{y\to\infty}$ of this expression, which I haven't done manually yet, but it looks like a doable task.
