Answer
$$-\frac{1}{2} (\ln( \cos^{-1}x))^2+c.$$
Work Step by Step
Let $u=\ln( \cos^{-1}x)$, then $du =-\frac{dx}{(\cos^{1-}x)\sqrt{1-x^2}}$. Now, we have
$$\int \frac{\ln( \cos^{-1}x)dx}{(\cos^{1-}x)\sqrt{1-x^2}}=-\int udu\\
=-\frac{1}{2} u^2+c=-\frac{1}{2} (\ln( \cos^{-1}x))^2+c.$$
