Seemingly Simple Integral $\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx$. Evaluate $$\int_0^1 f(x) dx$$ where
$$f(x) = \frac{x^2\ln x}{\sqrt{1-x^2}}$$
I started off with the substitution $x=\sin y$, which resulted in the integrand reducing to
$$\sin^2y\cdot \ln (\sin y) dy$$
Then I used the property of definite integrals that
$$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$
Then too it wasn't getting simplified.
I tried $e^z=\sin x$, but this gave no headway because after a while I reached a complete full-stop. How should I go about this?

$$
\begin{aligned}
I &\stackrel{x=\sin \theta}{=} \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta \ln (\sin \theta)}{\cos \theta} \cdot \cos \theta d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \sin ^{2} \theta \ln (\sin \theta) d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right)\\& \stackrel{IBP}{=}  \left[\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right) \ln (\sin \theta)\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right)d(\ln(\sin \theta) )\\& =-\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\theta}{2} d(\ln (\sin \theta))}_{J}+\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\sin 2 \theta}{4} d(\ln (\sin \theta))}_{K}
\end{aligned}
$$
Integration by parts yields
$$\begin{aligned} J &=\left[\frac{\theta}{2} \ln (\sin \theta)\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d \theta =\frac{\pi}{4} \ln 2 \end{aligned}$$
$$
\begin{aligned}
K &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin 2 \theta}{\sin \theta} \cos \theta d \theta =\frac{1}{4} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 \theta) d \theta =\frac{\pi}{8} \\
\end{aligned}
$$
Now we can conclude that
$$\boxed{I =\frac{\pi}{8}(1-\ln 4)}$$
