Integration by parts yields divergent integral 
$$I = \int_0^1\dfrac{t \ln t}{\sqrt{1-t^2}}\mathrm{dt}$$

Attempt: 
Let $t = \sin x$
$\implies dt = \cos x dx$
$\implies I = \displaystyle \int_0^{\pi/2} \dfrac{\sin x \ln (\sin x) \cos x dx}{\cos x}$
$\implies  I = -\ln (\sin x) \cos x|_{0}^{\pi/2}   + \displaystyle\int_0^{\pi/2}\dfrac{\cos^2 x}{\sin x} dx$
The second integral can be "evaluated"  using $\cos^2 x = 1- \sin^2 x$. But unfortunately, there are finally two divergent integrals. 
$\implies  I = -\ln (\sin x) \cos x|_{0}^{\pi/2}   + \displaystyle\int_0^{\pi/2}\csc  x dx$ + $\displaystyle\int_0^{\pi/2}\sin x dx$
How do I handle this? 
 A: If you want to avoid divergent integrals, use $1-\cos x$ as the antiderivative of $\sin x$.
$$I=\ln\sin x(1-\cos x)\Big|_0^{\pi/2}+\int_0^{\pi/2}(\cos x-1)\cot xdx\\
=0+\ln2-1=\ln2-1.$$
A: I would just focus on the antiderivative.
Using you substitution and continuing with the tangent half-angle substitution, we should get
$$I=\int \dfrac{\sin (x) \ln (\sin (x)) \cos(x) }{\cos x}\,dx=\int\sin (x) \ln (\sin (x))\,dx$$ and integration by parts would lead to
$$I=\cos (x)+\log \left(\sin \left(\frac{x}{2}\right)\right)-\log \left(\cos
   \left(\frac{x}{2}\right)\right)-\cos (x) \log (\sin (x))$$
Now, looking at the Taylor expansion around $x=0$,
$$I=(1-\log (2))+x^2 \left(\frac{\log (x)}{2}-\frac{1}{4}\right)+O\left(x^4\right)$$ and around $x=\frac \pi 2$
$$I=-\frac{1}{6} \left(x-\frac{\pi }{2}\right)^3+O\left(\left(x-\frac{\pi
   }{2}\right)^4\right)$$ and then the result already given by @Kemono Chen.
Edit
Back to $t$,
$$\int\dfrac{t \ln t}{\sqrt{1-t^2}}\,{dt}=\sqrt{1-t^2} (1-\log (t))+\log \left(\frac{t}{1+\sqrt{1-t^2}}\right)$$
