I have to show that the following the integral in terms of $x$ can be rewritten as follows:
$$\int_{-1}^1\frac{\sqrt{1-x^2}}{1+x^2}\ \mathrm dx = \int_{u_2=}^{u_1=}\frac1{1+cos^2u}\ \mathrm du-\pi$$
I have used the substitution $x=\cos u$
My working is as follows, I'm just struggling to manage to get the $-\pi$ term:
$$\begin{align} &\int_{\pi}^{0}\frac{\sqrt{1-\cos^2u}}{1+\cos^2u}\cdot-\sin u\ \mathrm du\\[5pt] &\int_{\pi}^{=0}\frac{-\sin^2 u}{1+\cos^2u}\ \mathrm du\\[5pt] &\int_{0}^{\pi}\frac{\sin^2 u}{1+\cos^2u}\ \mathrm du \end{align}$$