Integral $ \int_{0}^{\pi/2} \frac{\pi^{(x^{e})}\sin(x)\tan^{-1}(x)}{\sinh^{-1}\left({1+\cos(x)}\right)} dx$

Question

I need help in evaluating the following integral :-

$$ \int_{0}^{\pi/2} \frac{\pi^{\displaystyle (x^{e})}\sin(x)\tan^{-1}(x)}{\sinh^{-1}\left({1+\cos(x)}\right)} dx$$

A brief solution would be very much appreciated.

Answer 1

By using the formula http://upload.wikimedia.org/wikipedia/en/math/7/2/a/72a1058ad2087aec467af24bddcf9479.png, we have $\int_0^\frac{\pi}{2}\dfrac{\pi^{x^e}\sin x\tan^{-1}x}{\sinh^{-1}(1+\cos x)} dx=\dfrac{\pi}{2}\sum\limits_{n=1}^\infty\sum\limits_{m=1}^{2^n-1}\dfrac{(-1)^{m+1}\pi^\frac{m^e\pi^e}{2^{e(n+1)}}\sin\dfrac{m\pi}{2^{n+1}}\tan^{-1}\dfrac{m\pi}{2^{n+1}}}{2^n\sinh^{-1}\left(1+\cos\dfrac{m\pi}{2^{n+1}}\right)}$

Answer 2

Instead of you I would first check that the integrand is defined on ]0,π[, and if so, that it is integrable. You may have a surprise ! ;-)

I Remember having long time ago such an exercise where the integrand wasn't even defined in the middle of the interval, and the exercise was given to us only to check if we were checking the basic stuff before trying to calculate the integral.