How do you set up this Tricky u-sub?

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: $$\int{{x^3}\sqrt{5-2x^2}}\,\mbox{d}x = -\frac{1}{4}\int\frac{5-u}{2}\sqrt{u}\,\mbox{d}u= -\frac{1}{8}\int \left(5\sqrt{u}-u^{\tfrac{3}{2}}\right)\,\mbox{d}u = \ldots$$

By your idea we have

• $u = 5-2x^2\implies x^2=\frac{5-u}{2}$
• $ du = -4x dx\implies xdx=-\frac14 du $

then

$$\int{{x^3}\sqrt{5-2x^2}}dx=\int{\frac{5-u}{2}\sqrt{u}}\left(-\frac14\right) du$$