# 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? • In a later comment, you mention that your answer doesn't match the book's. The only thing you seem to be missing from your work is that you can't have an integral of x end up being about u. You have to change variables back to the original x. May 8, 2018 at 15:26

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

• Ok, that's exactly where I got to. Now, I just integrate the basic binomial, right? I edited the original post with my work. This did not match the solutions in the book I am working with. Theirs was all a very simple integral with a single term. May 8, 2018 at 15:01
• @JackOfAll That would have been something to include from the beginning. May 8, 2018 at 15:24
• @JackOfAll The steps you added are all fine! Now replace $u$ by $5-2x^2$ and you are done. If the answer doesn't look the same; they may have simplified a bit further. May 8, 2018 at 16:16
• The answer in the book was nothing like mine. They reduced the entire integral to something like $-\frac{1}{4}\int{u^\frac{2}{3}du}$ May 8, 2018 at 16:25
• @JackOfAll - Unless you supply more details, we cannot guess what is going on with your "something like"s. I strongly suspect you are mis-interpreting what the book is saying. But again, without details, I can't be sure. May 8, 2018 at 16:53

• $u = 5-2x^2\implies x^2=\frac{5-u}{2}$
• $du = -4x dx\implies xdx=-\frac14 du$
$$\int{{x^3}\sqrt{5-2x^2}}dx=\int{\frac{5-u}{2}\sqrt{u}}\left(-\frac14\right) du$$
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 $u$.