Integrate ∫π03cosx+√8+cos2xsinxx dx\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx

Please help me to solve this integral:
π03cosx+8+cos2xsinxx dx.

I managed to calculate an indefinite integral of the left part:
cosxsinxx dx= xlog(2sinx)+12 Li2(e2 x i),
where  Li2(z) denotes the imaginary part of the dilogarithm. The corresponding definite integral π0cosxsinxx 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
π03cosx+8+cos2xsinxx dx?=πlog54,
but I have no idea how to prove it.

Answer

Here’s one way to go.

First, note that
π03cosx+8+cos2xsinxx dx=π03x(1+cosx)sinxdx+π03xsinx(1+1sin2x9) dx.
For now I’ll simply claim that
π03x(1+cosx)sinxdx=πlog64.
(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*}

Attribution
Source : Link , Question Author : Piotr Shatalin , Answer Author : user26872

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