Closed form of $\int_{0}^{\eta}\cos nt\log\left(\frac{\cos(t/2)+\sqrt{\cos^2(t/2) -\cos^2(\eta/2)}}{\cos(\eta/2)}\right) dt$ I am reading a paper (sorry, no e-copy) with a number of infinite series, in which each term of the series is an integral of a complicated transcendental function like the one in the title. 
There are no fewer than a dozen of these infinite sums of integrals in the paper.
For instance, the authors define
$$
I_1(\eta) = \sum_{n=2,3,4}^{\infty} \dfrac{n}{n^2-1} \int_{0}^{\eta} \cos nt \log \left( \dfrac{\cos(t/2) + \sqrt{\cos^2(t/2) - \cos^2(\eta/2)}}{\cos(\eta/2)}\right) dt
$$ 
where $0 \leq \eta < \pi$. The authors then assert that
$$
I_1(\eta) = \frac{\pi}{8} \left( (2\cos\eta -2) \log \sin \frac{\eta}{2}  + \frac{1}{2}\cos\eta -\frac{1}{2}\right)
$$
without any proof or derivation (the results are simply listed in an Appendix). 
I confess I am stumped as to how the authors arrived at their results. Any advice or help would be greatly appreciated. 
 A: $\newcommand{\s}{\sigma}$$\newcommand{\e}{\eta}$$\newcommand{\al}[1]{\begin{align}#1\end{align} }$As suggested by Norbert Schuch in the comments, it suffices to check that the derivatives with respect to $\eta$ agree. This simplifies our task somewhat by getting rid of one of the logarithms in the integrand.
Starting from the expression derived by JJacquelin, we have for the case $0 \leq \e < \pi/2$:
$$
\al{
I_1'(\e) &= -\frac{\tan \frac \e 2}{8} \int_0^\e d\e \frac{ \cos \frac{t}{2} \left[2+ \cos t+2 \cos t \ln (2-2\cos t)\right]}{ \sqrt{\cos^2 \frac t 2 - \cos^2 \frac \e 2}}
\\&= -\frac{\tan \frac \e 2}{8} \int_0^\e d\e\frac{ \cos \frac{t}{2} \left[2 + (1 - 2 \sin^2 \frac t 2)+2 (1 - 2 \sin^2 \frac t 2) \ln (4 \sin^2 \frac t 2)\right]}{ \sqrt{\sin^2 \frac \e 2 - \sin^2 \frac t 2}}
\\&= -\frac{\tan \frac \e 2}{8} \int_0^1 dx\frac{ 2\left[3 - 2 x^2 \sin^2 \frac \e 2+2 (1 - 2 x^2 \sin^2 \frac \e 2) \ln (4 x^2 \sin^2 \frac \e 2)\right]}{ \sqrt{1 - x^2}}
\\&\equiv -\frac{\tan \frac \e 2}{4} I
}
$$
Here I put $\sin \frac t 2 = x \sin \frac \e 2 $, and there is no boundary term because the integrand vanishes at $t = \e$. 
Now we abbreviate $\s = \sin \frac \e 2$ for clarity. We have
$$
I = A+B+C+D+E+F,
$$
with 
$$
\al{
&A=\int_0^1 dx \frac{3}{\sqrt{1-x^2}}=\frac{3\pi}{2}
\\&B=\int_0^1 dx \frac{-2 \s^2 x^2}{\sqrt{1-x^2}}= -\frac \pi 2 \s^2
\\&C=\int_0^1 dx \frac{2 \ln (4\s^2)}{\sqrt{1-x^2}}= \pi \ln (4\s^2)
\\&D=\int_0^1 dx \frac{4 \ln x}{\sqrt{1-x^2}}= 4 \int_0^{\pi/2} dt \ln \sin t = -\pi \ln 4
\\&E=\int_0^1 dx \frac{-4x^2 \s^2 \ln(4\s^2)}{\sqrt{1-x^2}}= -\pi \s^2 \ln (4\s^2)
\\&F=\int_0^1 dx \frac{-4x^2 \s^2 (2 \ln x)}{\sqrt{1-x^2}}= -8 \s^2 \int_0^{\pi/2} dt \sin^2 t\ln \sin t = \pi \s^2 (\ln 4-1).
}
$$
Therefore,
$$
I = \pi(1-\s^2) \left[ \frac{3}{2} + 2 \ln \s \right] = \pi \cos^2 \frac \e 2 \left[ \frac{3}{2} + 2 \ln \sin \frac \e 2\right].
$$
This gives
$$
I_1'(\e) = -\frac \pi 4 \sin \frac \e 2 \cos \frac \e 2 \left[ \frac{3}{2} + 2 \ln \sin \frac \e 2\right] = -\frac \pi 8 \sin \e \left[ \frac{3}{2} + 2 \ln \sin \frac \e 2\right].
$$
This agrees with the derivative of the desired result. 
The case $\pi/2 \leq \eta \leq \pi$ can be handled similarly; the only thing that changes is that now we cannot assume that $\cos^2 \frac t 2 > \cos^2 \frac \e 2$ everywhere.
A: Numerical tests show a very good agreement between the two forms of $I_1(\eta)$ So, I am conviced that the assertion of the authors is correct.
By the way, it is possible to get ride of the discrete sum, thanks to :
$$
\sum_{n=2}^{\infty} \dfrac{n}{n^2-1}\cos(nt)=-\frac{1}{4}\left(2+\cos(t)+2\cos(t)\ln \left( 2-2\cos(t)\right) \right)
$$ 
leading to :
$$
I_1(\eta) = -\int_{0}^{\eta}\frac{2+\cos(t)+2\cos(t)\ln \left( 2-2\cos(t)\right)}{4}
 \ln \left( \dfrac{\cos(\frac{t}{2}) + \sqrt{\cos^2(\frac{t}{2})  - \cos^2(\frac{\eta}{2})}}{\cos(\frac{\eta}{2})}\right) dt
$$
Not sure that it will be useful.
