$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}}$
Can anyone give me a hint where to go with this?
Another cool way to do this is let $\displaystyle u=1-x^2$, (which also implies that $\displaystyle x^2=1-u$)
Then $\displaystyle du=-2xdx$
You can then substitute $\displaystyle 1-x^2$ with $\displaystyle u$, $\displaystyle x^2$ with $\displaystyle 1-u$, the "other" $\displaystyle x$ in the denominator with $\displaystyle du$ (as long you put the $\displaystyle -\frac{1}{2}$ out front.) Then the integral becomes $\displaystyle -\frac{1}{2}\int \frac{1-u}{u^\frac{1}{2}}du$, which can then be easily evauluated by bringing up the $\displaystyle u^\frac{1}{2}$ to the denominator