Is brute force the only answer to $\int u^2(1-u^2)^4du$? 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: $\int \sin^2x\cos^9x\;dx$
 A: 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.
A: Disclaimer: This answer is only partially serious; you'd have to have Ramanujan's affection to numbers in order to to proceed like this :-)
So, here's the one-liner solution which doesn't actually require you to perform any integration at all (well, split into multiple lines for the sake of readability):
$$\begin{array}{ll}
\int f(x)\ \mathrm{d}x=\frac{x}{598752}\Big[
& 16067\left(f\left(\frac{0}{10}x\right)+f\left(\frac{10}{10}x\right)\right) \\
& +106300\left(f\left(\frac{1}{10}x\right)+f\left(\frac{9}{10}x\right)\right) \\
& -48525\left(f\left(\frac{2}{10}x\right)+f\left(\frac{8}{10}x\right)\right) \\
& +272400\left(f\left(\frac{3}{10}x\right)+f\left(\frac{7}{10}x\right)\right) \\
& -260550\left(f\left(\frac{4}{10}x\right)+f\left(\frac{6}{10}x\right)\right) \\
& +213684\left(f\left(\frac{5}{10}x\right)+f\left(\frac{5}{10}x\right)\right) \ \Big] + C
\end{array}
$$
In fact, we don't even need to know what $f(x)$ is; as long as we are able to evaluate it in the eleven points (one of them appears twice in the expresion) and know that it's a polynomial of degree at most ten! The magic comes from the $11$-point Newton-Cotes formula.

And if you notice the integrand is actually a function of $u^2$ instead of just $u$, you can even reduce the number of evaluations to just six (the coefficients remained the same; just the evaluated arguments changed a bit):
$$
\int f(x)\ \mathrm{d}x=\frac{2x}{598752}\left[
16067 f\left(\frac{5}{5}x\right) +106300 f\left(\frac{4}{5}x\right) -48525 f\left(\frac{3}{5}x\right) +272400 f\left(\frac{2}{5}x\right) -260550 f\left(\frac{1}{5}x\right) + 213684 f\left(\frac{0}{5}x\right) \right] + C
$$
A: I don't know if this is what you want, but we can express the integral in terms of the Incomplete Beta Function. But first you must notice that (when $C=0$),
$$\int x^2(1-x^2)^4dx=\int_0^x t^2(1-t^2)^4dt$$ 
Then we look at the definite integral 
$$I(x)=\int_0^x t^2(1-t^2)^4dt$$
And make the substitution $u=t^2$, which gives
$$I(x)=\int_0^{x^2} u(1-u)^4\frac12u^{-1/2}du$$
$$I(x)=\frac12\int_0^{x^2}u^{-1/2}(1-u)^4du$$
$$I(x)=\frac12\int_0^{x^2}u^{-1/2}(1-u)^4du$$
$$I(x)=\frac12\int_0^{x^2}u^{1/2-1}(1-u)^{5-1}du$$
Then we recall the definition of the incomplete Beta function
$$B(x;a,b)=\int_0^xt^{a-1}(1-t)^{b-1}dt$$
This gives our integral (add $C$ if you want)
$$I(x)=\frac12B\bigg(x^2;\frac12,5\bigg)$$
Interesting Special Value:
By noting that ($\Gamma(s)$ is the Gamma Function)
$$B(1;a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
We have 
$$I(1)=\frac{\Gamma(\frac12)\Gamma(5)}{2\Gamma(\frac{11}2)}$$
$$I(1)=\frac{24\sqrt{\pi}}{9\Gamma(\frac{9}2)}$$
$$I(1)=\frac{8\sqrt{\pi}}{3\cdot\frac72\Gamma(\frac{7}2)}$$
Blah blah blah (the sound of numerical simplification)
$$I(1)=\frac{256}{315}$$
