Compute $\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}$ Compute $$\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}.$$
I started by making the substitution $\arccos x = t$. Hence, $-\frac{dx}{\sqrt{1-x^2}}=dt$. 
Now I get that $$\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}=-\int\limits_{\pi/3}^{2 \pi/3}\frac{dt}{t^2}=\frac{3}{2\pi}-\frac{3}{\pi}=-\frac{3}{2\pi}.$$ 
However, the result should be $\dfrac{3}{2\pi}$. What did I do wrong?

$\arccos(-0.5)=\frac{2\pi}{3},\arccos(0.5)=\frac{\pi}{3}$ thus transformed integral must be $-\int_{\frac{2\pi}{3}}^{\frac{\pi}{3}}(...)$
