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?
 A: Consider the integral $$I\left(t\right)=\frac{1}{2}\int_{0}^{1}\frac{x^{2t}}{\sqrt{1-x^{2}}}dx$$ we can observe that $$I'\left(1\right)=\int_{0}^{1}\frac{x^{2}\log\left(x\right)}{\sqrt{1-x^{2}}}dx.$$ So we have $$I\left(t\right)=\frac{1}{2}\int_{0}^{1}\frac{x^{2t}}{\sqrt{1-x^{2}}}dx\stackrel{x^{2}=v}{=}\frac{1}{4}\int_{0}^{1}\frac{v^{t-1/2}}{\sqrt{1-v}}dv$$ and using the identity involving the hypergeometric function $$_{2}F_{1}\left(a,b;c;z\right)=\frac{\Gamma\left(c\right)}{\Gamma\left(b\right)\Gamma\left(c-b\right)}\int_{0}^{1}\frac{v^{b-1}\left(1-v\right)^{c-b-1}}{\left(1-vz\right)^{a}}$$ under the hypothesis $$\textrm{Re}\left(c\right)>\textrm{Re}\left(b\right)>0\wedge\left|\textrm{arg}(1-z)\right|<\pi$$ we get $$I\left(t\right)=\frac{\Gamma\left(t+1/2\right)}{4\Gamma\left(t+3/2\right)}{}_{2}F_{1}\left(\frac{1}{2},t+\frac{1}{2};t+\frac{3}{2};1\right)$$ and again using the closed form $$_{2}F_{1}\left(a,b;c;1\right)=\frac{\Gamma\left(c\right)\Gamma\left(c-a-b\right)}{\Gamma\left(c-a\right)\Gamma\left(c-b\right)},\,\textrm{Re}\left(c-a-b\right)>0$$ we have $$I\left(t\right)=\frac{\Gamma\left(t+1/2\right)}{4\Gamma\left(t+3/2\right)}\frac{\Gamma\left(t+3/2\right)\Gamma\left(1/2\right)}{\Gamma\left(t+1\right)\Gamma\left(1\right)}=\frac{\Gamma\left(1/2\right)}{4}\frac{\Gamma\left(t+1/2\right)}{\Gamma\left(t+1\right)}$$ so now we can take the derivative and evalutate it at $t=1$ $$I'\left(1\right)=\frac{\Gamma\left(1/2\right)}{4}\frac{\Gamma\left(3/2\right)\left(\psi\left(3/2\right)-\psi\left(2\right)\right)}{\Gamma\left(2\right)}=-\frac{1}{8}\pi\left(\log\left(4\right)-1\right)\approx-0.151697$$ where $\psi(x)$ is the digamma function.
A: An answer that does not use hypergeometric functions:
First integrate by parts using the functions $u(x)=x\ln x$ and $v(x)=-\sqrt{1-x^2}$, we have $u'(x)=\ln x +1$ and $v'(x)=\frac{x}{\sqrt{1-x^2}}$ and we get
$$I=\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}\mathrm dx=
\underbrace{\left[-x\ln x\sqrt{1-x^2}\right]_0^1}_{=0}+\int_0^1(\ln x+1)\sqrt{1-x^2}\,\mathrm dx.$$
The integral $\int_0^1\sqrt{1-x^2}\mathrm dx=\frac\pi4$ is easy. Let us concentrate on 
$$J=\int_0^1\ln x \,\sqrt{1-x^2}\mathrm dx$$
for which it seems a trigonometric change of variable will work. Let us set $x=\cos\theta$, $\mathrm dx=-\sin\theta\mathrm d\theta$, then
$$J=\int_0^{\pi/2}\ln(\cos\theta)\sin^2\theta\mathrm d\theta
=\int_0^{\pi/2}\ln(\cos\theta)(1-\cos^2\theta)\mathrm d\theta.\tag1$$
We can use the result 
$$\int_0^1\ln(\sin\theta)\,\mathrm d\theta=\int_0^{\pi/2}\ln(\cos\theta)\,\mathrm d\theta=-\frac\pi2\ln2$$
(see for instance this post for a derivation). Therefore we have 
$$J=-\frac\pi2\ln2-\int_0^{\pi/2}\ln(\cos\theta)\cos^2\theta\,\mathrm d\theta.\tag2$$
Adding up (1) and (2) we obtain
$$2J=-\frac\pi2\ln2+\int_0^{\pi/2}\ln(\cos\theta)\left(\sin^2\theta-\cos^2\theta\right)\,\mathrm d\theta=-\frac\pi2\ln2-\int_0^{\pi/2}\ln(\cos\theta)\cos(2\theta)\,\mathrm d\theta.$$
Finally let us integrate by parts (with $u(x)=\ln(\cos\theta)$ and $v(x)=\frac12\sin(2\theta)$, $u'(x)=-\tan\theta$ and $v'(x)=\cos(2\theta)$)
$$\begin{split}\int_0^{\pi/2}\ln(\cos\theta)\cos(2\theta)\,\mathrm d\theta&=
\underbrace{\left[\ln(\cos\theta)\frac12\sin(2\theta)\right]_0^{\pi/2}}_{=0}
+\int_0^{\pi/2}\tan\theta\,\frac12\sin2\theta\mathrm d\theta\\
&=\int_0^{\pi/2}\sin^2\theta\,\mathrm d\theta=\frac\pi4
\end{split}$$
We get $2J=-\frac\pi2\ln2-\frac\pi4$ and therefore
$$I=-\frac\pi4\ln2-\frac\pi8+\frac\pi4=\boxed{\frac\pi8-\frac\pi4\ln2}.$$
A: $$I(\alpha)=\int_0^1\dfrac{x^\alpha\ln x}{\sqrt{1-x^2}}\ dx$$
Using Beta function we have
$$\int_0^1\dfrac{x^\alpha}{\sqrt{1-x^2}}\ dx=\dfrac12{\bf B}\left(\frac{\alpha+1}{2},\frac{1}{2}\right)$$
then with Digamma function $\psi$
\begin{align}
I(\alpha)
= \dfrac{d}{d\alpha}\int_0^1\dfrac{x^\alpha}{\sqrt{1-x^2}}\ dx \\
= \dfrac12{\bf B}\left(\frac{\alpha+1}{2},\frac{1}{2}\right)\left(\psi(\frac{\alpha+1}{2})-\psi(\frac{\alpha+2}{2})\right)
\end{align}
now let $\alpha=2$,
$$\psi(\frac{3}{2})-\psi(\frac{4}{2})=2\sum_{n=0}^\infty\left(\dfrac{1}{2n+4}-\dfrac{1}{2n+3}\right)=1-\ln4$$
therefore
$$I(2)=\color{blue}{\dfrac{\pi}{8}(1-\ln4)}$$
A: $$
\begin{aligned}
I &\stackrel{x=\sin \theta}{=} \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta \ln (\sin \theta)}{\cos \theta} \cdot \cos \theta d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \sin ^{2} \theta \ln (\sin \theta) d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right)\\& \stackrel{IBP}{=}  \left[\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right) \ln (\sin \theta)\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right)d(\ln(\sin \theta) )\\& =-\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\theta}{2} d(\ln (\sin \theta))}_{J}+\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\sin 2 \theta}{4} d(\ln (\sin \theta))}_{K}
\end{aligned}
$$
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
$$\boxed{I =\frac{\pi}{8}(1-\ln 4)}$$
A: Noting
$$ d\bigg[-\frac12x\sqrt{1-x^2}+\frac12\arcsin(x)\bigg]=\frac{x^2}{\sqrt{1-x^2}}, \int_0^{\pi/2}\ln\sin xdx=-\frac{\pi}{2}\ln2$$
one has
\begin{eqnarray}
\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx&=&\int_0^1\ln xd\bigg[-\frac12x\sqrt{1-x^2}+\frac12\arcsin(x)\bigg]\\
&=&\bigg[-\frac12x\sqrt{1-x^2}+\frac12\arcsin(x)\bigg]\ln x\bigg|_0^1-\int_0^1\bigg[-\frac12x\sqrt{1-x^2}+\frac12\arcsin(x)\bigg]\frac1xdx\\
&=&\frac12\int_0^1\sqrt{1-x^2}dx-\frac12\int_0^1\frac{\arcsin x}{x}dx\\
&=&\frac{\pi}{8}-\frac12\int_0^{\pi/2}\frac{x}{\sin x}\cos xdx\\
&=&\frac{\pi}{8}-\frac12\int_0^{\pi/2}xd\ln\sin x\\
&=&\frac{\pi}{8}-\frac12x\ln\sin x\bigg|_0^{\pi/2}+\frac12\int_0^{\pi/2}\ln\sin xdx\\
&=&\frac{\pi}{8}-\frac\pi4\ln 2.
\end{eqnarray}
