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What's the solution of this integral? $$\int_{0}^{\pi} \frac{\cos^2 \left( \dfrac{\pi \cos x}{2} \right)} {\sin x} \, dx$$

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    It has no elementary solution unless you consider [trigonometric integrals](https://en.wikipedia.org/wiki/Trigonometric_integral) as elementary. – 2017-01-16
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    It would help to know more about how far you got with your thinking on this problem, and what kind of result (numeric vs. symbolic) you hope to achieve. In particular the apparent singularity in the denominator at endpoints $x= 0,\pi$ is compensated by the behavior of the numerator at these points. – 2017-01-16

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It is easy to see that $~I~=~2\displaystyle\int_0^1\frac{\cos^2\Big(\dfrac\pi2~x\Big)}{1-x^2}~dx~=~\dfrac4\pi\displaystyle\int_0^\tfrac\pi2\frac{\cos^2x}{1-(ax)^2}~dx,~$ for $~a=\dfrac2\pi~.$

Judging by its initial integral expression, it would appear that I is connected to the topic of

Bessel functions. Judging by the latter, however, its link to trigonometric integrals becomes

self-evident. Indeed, on one hand we have $~I~=~\dfrac{\gamma+\ln(2\pi)-\text{Ci}(2\pi)}2,~$ while on the other

we get $~I~=~\dfrac{\gamma+\ln(2\pi)}2~+~\dfrac\pi4\sqrt2\cdot J^{(1,0)}\bigg(-\dfrac12~,~\pi\bigg).~$ By comparing the two results, we

ultimately arrive at the conclusion that $~J^{(1,0)}\bigg(-\dfrac12~,~\pi\bigg)~=~-\dfrac{\sqrt2}\pi~\text{Ci}(2\pi).$

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    In case you're wondering, $\gamma$ is the [Euler-Mascheroni constant](http://en.wikipedia.org/wiki/Euler-Mascheroni_constant), and its connection to [trigonometric integrals](http://en.wikipedia.org/wiki/Trigonometric_integral) is given by [this](http://math.stackexchange.com/questions/549587) auspicious result. – 2017-01-16