$$ \begin{align} \int \cos^{-1} x \; dx &= \int \cos^{-1} x \times 1 \; dx \end{align} $$
Then, setting $$\begin{array}{l l} u=\cos^{-1} x & v=x \\ u' = -\frac{1}{\sqrt{1-x^2}} & v'=1\\ \end{array}$$
Then by the IBP technique, we have:
$$\begin{array}{l l} \int \cos^{-1} x \times 1 \; dx &=\cos^{-1} (x) \cdot x - \int x \cdot -\frac{1}{\sqrt{1-x^2}} \; dx\\ &= x \cos^{-1} (x) - \int -\frac{x}{\sqrt{1-x^2}} \; dx\\ \end{array}$$
Now at this point suppose I have overlooked the possibility of using integration by substitution (setting $u=1-x^2$) to simplify the second integral. Instead, I attempt to reapply IBP to the second integral $\int -\frac{x}{\sqrt{1-x^2}} \; dx$.
I let $$\begin{array}{l l} u= -\frac{1}{\sqrt{1-x^2}} = -(1-x^2)^{-\frac{1}{2}} \qquad & v= \frac{x^2}{2} \\ u' = - \left( -\frac{1}{2} \right) (1-x^2)^{-\frac{3}{2}} \times -2x = -x(1-x^2)^{-\frac{3}{2}} \qquad & v'=x\\ \end{array}$$
Then by IBP again,
$$\begin{array}{l l} \int \cos^{-1} x \times 1 \; dx &= x \cos^{-1} (x) - \left( -\frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot -x(1-x^2)^{-\frac{3}{2}} \; dx \right) \\ \end{array}$$
At this stage, I can see no way to proceed. Can anyone see a reasonable way to salvage this solution, continuing along this line of reasoning? Or was approaching the second integral by IBP doomed to fail?