Evaluate $\int\limits_{-1/8}^{1/8}\arccos\left(x^2+5x\right)\ dx$

Question

Find the value of $$\int\limits_{-1/8}^{1/8}\arccos\left(x^2+5x\right)\ dx$$

Answer given

$\dfrac{\pi}{8}$

I used property and wrote the integral as $$\dfrac{1}{2}\int\limits_{-1/8}^{1/8} \left( \arccos\left(x^2+5x\right) + \arccos\left(x^2-5x\right) \right)\ dx$$, then added two inverse functions. But didn't prove to be useful. How should I do it?

Answer 1

I've been wrong before but I'm not sure I buy the answer is $\pi/8$. Using the complementary angle formula, we have $$ \int _{-1/8}^{1/8}\arccos(x^2+5x)\,dx = \int _{-1/8}^{1/8}\frac{\pi}{2}-\arcsin(x^2+5x)\,dx $$ $$ = \frac{\pi}{8}- \int _{-1/8}^{1/8}\arcsin(x^2+5x)\,dx $$But I don't think this last integral is $0$. The range is symmetric and $\arcsin$ is odd, but the argument isn't odd; it's slightly positive on $[-1/8,1/8]$ by a derivative argument. Numerical calculation suggests the answer isn't $\pi/8$ either.