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?

Setting $\arccos x = t$ gives us $\frac{-dx}{\sqrt{1-x^2}} = dt$
So our integral is then: 
$$\int_{2\pi/3}^{\pi/3} \frac{-dt}{t^2}$$
$$=\int_{\pi/3}^{2\pi/3} \frac{dt}{t^2}$$
$$= \frac{3}{\pi}-\frac{3}{2\pi}$$
$$=\frac{3}{2\pi}$$
