I have the integral $\int_{-1}^{1} \frac{\arccos(x)}{1+x^2} \,dx $ and some questions. Any help appreciated!

Question

$$\int_{-1}^{1} \frac{\arccos(x)}{1+x^2} \,dx $$

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!

Answer 1

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 $$ I(1) = \int_{0}^{1} I'(a) \ \mathrm{d}a +I(0) = I(0) = \frac{\pi}{2}\int_{-1}^{1}\frac{1}{1+x^2}\ \mathrm{d}x = \boxed{\frac{\pi^2}{4}} $$

Answer 2

Wow, I put this problem for a high school integration bee. Here's my fastest solution!!

$$\int_{-1}^{1} \frac{\cos^{-1}(x)+(\sin^{-1}(x)-\sin^{-1}(x))}{x^2+1}dx$$

Note that $\cos^{-1}(x)+\sin^{-1}(x)=\frac{\pi}{2}$ and arcsinx is an odd function, so by symmetry the leftover arcsinx term is 0.

$$\int_{-1}^{1} \frac{\frac{\pi}{2}}{x^2+1}dx=\boxed{\frac{\pi^2}{4}}$$

Answer 3

By using king property :

$$\int_{-1}^{1} \frac{\arccos(x)}{1+x^2} \mathrm{d}x = \int_{-1}^{1} \frac{\arccos(-x)}{1+x^2} \mathrm{d}x $$.

So :

$$2I = \int_{-1}^{1} \frac{\arccos(x)+\arccos(-x)}{1+x^2} \mathrm{d}x = \int_{-1}^{1} \frac{\pi}{1+x^2} $$.

Consequently :

$$I = \frac{\pi}{2} \int_{-1}^{1} \frac{1}{1+x^2} = \frac{\pi}{2} \left[ \arctan(x) \right]_{-1}^{1} = \frac{\pi^2}{4} = \frac{3\zeta(2)}{2} $$.