Feedback:
After some initial interest, it seems this question has been down voted. Jyrki Lahtonen may be right that spending time on unsolvable problems is pointless, but, to a student, at what point does it become clear that it's pointless?
I hadn't seen this particular integral before and neither have many others here I'm sure. Is it not instructive to at least consider it? And if it turns out, according to people who have experience with it, that the polynomial solution is the best one we have, then that can be placed as a definitive answer below, and even more useful it could include some instructive advice about integration techniques in general.
I'm lost for words as to why this would be considered a bad question, as a teacher I personally don't think it is. Students learning the various integration methods need to develop some instinct as to what's possible with those methods, why certain techniques work in certain situations and not others. What is it about this one in particular? At least it has a solution. At what point in your progress as a mathematician should you have developed the confidence to know when to stop looking in a particular direction?
Question:
Consider the integral $\int u^2(1-u^2)^4du$.
One way I know of doing this involves a trig substitution and then using a recursion formula on powers of $\cos x$. I don't want that solution because I already know about it.
Aside from that, is brute force, that is, just expanding the polynomial and integrating each power separately, the only other way of doing this, or am I just having a mental block on some neat way to solve this? I just need to be sure I'm not forgetting something.
(1) Brute force: $u^2(1-u^2)^4 = u^2 - 4u^4 + 6u^6 - 4u^8 + u^{10}$. Which, after all, is really only a couple of lines.
(2) Integration by parts: All the ideas I tried either got me back to where I started from, required more work than the brute force method, or led to trig substitution which has been covered. If there's a simple by parts solution, please, let's see it.
(3) Substitution example:
\begin{align*} v^2 = 1 - u^2 \Rightarrow u\;du = -v\;dv\\ \int u^2(1-u^2)\;du = -\int \sqrt{1-v^2}\;v^4\;dv \\ \ldots \text{ aargh } \ldots \end{align*}
(4) Original equivalent trig integral: $\int \sin^2x\cos^9x\;dx$
I think after all this time that the simple answer is yes. The use of the term brute force in my original question statement may be a bit of an exaggeration as
$$u^2(1-u^2)^4 = u^2 - 4u^4 + 6u^6 - 4u^8 + u^{10}$$
is not too brutal. It's just not an example of one of those integral solutions that are really elegant. Hopefully the extended discussion that has taken place and the extension of the OP prove to be instructive.