I have been trying to solve the following integral for a little while now
$I = \int_a^b \arccos\left( \frac{x}{\sqrt{(a+b)x - ab \,}\,}\right)\mathrm{d}x \quad 0
After doing some testing in maple, I was able to discover that $\displaystyle I=\frac{\pi}{4}\frac{(a-b)^2}{b+a} $
But I want to show this algebraically. (using mathematics)
I have done some effort and one can notice that the denominator equals $x^2-(x-a)(x-b)$. Another thing i tried using was the fact that
$\arccos(x) = \arcsin\left(\sqrt{1-x^2}\,\right)$
also if on tries solving it by parts
$ \left[ \arccos\left( \frac{x}{\sqrt{(a+b)x - ab \,}\,}\right) \right]_a^b = 0 $
but the remaining integral looks impossible. If one uses the substitution, $u=\text{denominator}$ I end up with
$ \frac{2}{a+b}\int u^2 \arccos\left( \frac{u^2 + ab}{u(b+a)} \right) \, \mathrm{d}u$
Also $I = \int_a^b \text{arccsc}\left( \sqrt{\frac{(a-x)(b-x)}{x^2}}\right)\mathrm{d}x$
not sure what this buys me though. I also tried a variety of clever things such at differentiating under the integral sign, but alas nothing worked. Anyone mind helping me ? =)