$$\int_{-1}^{1} \frac{\arccos(x)}{1+x^2} \,dx $$
Hi everyone! Sorry for my poor formatting skills, I'm still quite new to this platform.
I do not know how to solve this integral.
Things that I tried but failed miserably:
I've tried substituting $t = \frac{1}{1+x^2}$ but this (of course) does not work since its $\arccos(x)$ and not $\arctan(x)$.
I've tried solving it by parts but that also leads to a dead-end.
Since the bounds are $1$ and $-1$, I tried checking if the function is even or odd but $\arccos(x)$ is neither.
My questions (even though they might sound ridiculous or noobish) are:
Is there a Weierstrass-like substitution for inverse trigonometric functions? (tried googling it but to no avail)
How would one go about solving this type of integral?
Thanks in advance for help!
I offer another approach.
You can consider the integral $I(a) = \displaystyle{\int}_{-1}^{1}\dfrac{\arccos ax}{1+x^2}\ \mathrm{d}x$. Then $$ I'(a) = \int_{-1}^{1}\frac{\partial}{\partial a} \frac{\arccos ax}{1+x^2} \ \mathrm{d}x=-\int_{-1}^{1} \frac{x}{\sqrt{1-a^2 x^2}}\frac{1}{1+x^2} \ \mathrm{d}x=0 $$ The last step is because the integrand is odd. Finally, the integral can be calculated as $$ I(1) = \int_{0}^{1} I'(a) \ \mathrm{d}a +I(0) = I(0) = \frac{\pi}{2}\int_{-1}^{1}\frac{1}{1+x^2}\ \mathrm{d}x = \boxed{\frac{\pi^2}{4}} $$