How do you set up this Tricky u-sub? Tricky u-sub.  Can you point me in the right direction?
$$\int{{x^3}\sqrt{5-2x^2}}dx$$
$u = 5-2x^2$
$ du = -4x dx$
Obviously, this does not match fully.

I tried breaking up the $x^3 = x*x^2$
 and I continued with:
$u=5-2x^2$
$2x^2 = 5-u$
$x^2=\frac{5-u}{2}$
But, I can't see how to make it all fit.
Am I on the right track?

 A: 
Am I on the right track?

You're doing fine!
So you have $u=5-2x^2$ which gives $x^2=\frac{5-u}{2}$ and from $\mbox{d}u=-4x\,\mbox{d}x$ you get $x\,\mbox{d}x=-\tfrac{1}{4}\mbox{d}u$.
Now perform the substitution:
$$\int{{x^3}\sqrt{5-2x^2}}\,\mbox{d}x = -\frac{1}{4}\int\frac{5-u}{2}\sqrt{u}\,\mbox{d}u= -\frac{1}{8}\int \left(5\sqrt{u}-u^{\tfrac{3}{2}}\right)\,\mbox{d}u = \ldots$$
A: By your idea we have


*

*$u = 5-2x^2\implies x^2=\frac{5-u}{2}$

*$ du = -4x dx\implies xdx=-\frac14 du $


then
$$\int{{x^3}\sqrt{5-2x^2}}dx=\int{\frac{5-u}{2}\sqrt{u}}\left(-\frac14\right) du$$
A: You've done all the hard work. Doing some rearranging, you obtain
$$\int x^3 \sqrt{5-2x^2}\, dx = \int -\frac{1}{4} x^2 \sqrt{5-2x^2} \cdot (-4x)\, dx = \int -\frac{1}{4} \cdot \frac{5-u}{2} \sqrt{u}\, du$$
This last integral can be computed as usual by considering powers of $u$.
A: Another approach:
We can write the integrand as
\begin{align*}
{x^3}\sqrt{5-2x^2}&=-\tfrac12x(5-2x^2)\sqrt{5-2x^2}+\tfrac52x\sqrt{5-2x^2}&&\text{then}\\[5pt]
\int{x^3}\sqrt{5-2x^2}dx&=-\frac12\int x\left(5-2x^2\right)^{3/2}dx+\frac52\int x\left(5-2x^2\right)^{1/2}dx\\[5pt]
&=\color{red}{\frac1{20}\left(5-2x^2\right)-\frac5{12}\left(5-2x^2\right)^{3/2}+C}
\end{align*}
