\paragraph{Defining the interval on which $P_2(\eta) = P_{2_A}(\eta)$.} \\ Now since the integral is \textbf{not} partitioned into 2 regions when $-\frac{2\eta^2}{1 - \eta^2} = 0$ and when $-\frac{2\eta^2}{1 - \eta^2} = \eta$, we can determine the interval for which the root $s = -\frac{2\eta^2}{-1 + \eta^2}$ is located to the left of both $\eta$ and $1$ by looking for the solution to $$-\frac{2\eta^2}{1 - \eta^2} \leq \eta$$ on the interval $\eta \in [0,1]$