Having this:
$\int x\sqrt{1-x^2}dx$
Substitution:
$t = 1-x^2$
$dt = -2xdx => dx=\frac{-2x}{dt}$
So:
$$\int x\sqrt{1-x^2}dx = -\int x t^\frac{1}{2}\frac{2x}{dt} = -\int \frac{2x^2 t^\frac{1}{2}}{dt} = $$
...but here I stuck... I've tried $-4x\frac{t^\frac{-1}{2}}{\frac{-1}{2}} + c$ but it doesn't match correct result... why did I screw here?
You've mistakenly flipped your fraction for $dx$. The fact that you have a $dt$ in the denominator should tip you off, as that symbol doesn't have meaning on its own.
$$\int x\sqrt{1-x^2}dx=\dfrac{-1}{2}\int \sqrt{1-x^2}d(1-x^2)$$ Let $u=1-x^2$, and integrate. Note that if $-2xdx=dt$, then $xdx=\dfrac{-1}{2}dt$.
Use the fact that $\frac{d}{dx}((1-x^2)^\frac{3}{2})=-3x(1-x^2)^\frac{1}{2}$
(This is just using the rule $\frac{d}{dx}f(x)^n=f'(x)f(x)^{n-1}$)
And we have:
$\int x\sqrt{1-x^2}dx=(-1/3)\int 3x\sqrt{1-x^2}dx=(-1/3)\int \frac{d}{dx}((1-x^2)^\frac{3}{2})dx$
$=\int 1d((1-x^2)^\frac{3}{2})=-(1/3)(1-x^2)^\frac{3}{2}+c$
