Beginner Integration (Substitution) I am pretty new to calculus and would like a nudge in the right direction in order to complete this question properly (Maybe also correct any misrepresentations I have about integration)
So the question is this:
$$\int_{0}^{1}x\sqrt{1-\sqrt{x}}\:dx$$
My attempt at a solution:
let $$u=\sqrt{x}$$
$$u^2={x}$$
thus $$\frac{du}{dx}=\frac{1}{2}x^{-1/2}$$
$$dx=x^{1/2}2du=2udu$$
Now, $$\int_{a}^{b}u^{2}\sqrt{1-u}\:2udu\:$$
$$=2\int_{a}^{b}u^{2}\sqrt{1-u}\:udu$$
$$=2(\frac{1}{3}u^{3}*\frac{2}{3}({1-u})^{3/2}*\frac{1}{2}u^{2}*\frac{1}{2}u^{2})$$
I am able to substitute $\sqrt{x}$ for u, and solve the integral, but i don't believe this gives me the correct solution.
Can anyone point me towards a mistake, mis-conceptualization or different substitution that would allow me to solve this question? 
The question is tagged as dual substitution, does that mean i need to substitute two factors, or that multiple substitutions will work for this integral?
Thanks
P.S. Feel free to clean up my LaTex
 A: At the line $$=2\int_{0}^{1}u^{2}\sqrt{1-u}\:udu = 2\int_0^1 u^3\sqrt{1-u}\ du$$
(why did I set the bounds $=0$ and $1$?) you could do integration by parts, but since you say you're a "beginner", why not just use another substitution: Let $w=1-u$, then $dw=-du$.
$$=-2\int_1^0 (1-w)^3\sqrt{w}\ dw = 2\int_0^1 (1-w)^3\sqrt{w}\ dw$$ Then if you expand this out (meaning $(1-w)^3\sqrt{w} = (1 -3w + \cdots)\sqrt{w} = w^{1/2}-3w^{3/2}+\cdots$) you'll just be left with several integrals of the type $\int aw^k\ dw$ which hopefully you know how to evaluate.  Good luck. :)
A: In my previous answer I ended where I assumed you could take over.  I guess that was a false assumption.  In this answer I'll try to explain each step.
Start with the integral $$\int_{x=0}^{x=1} x\sqrt{1-\sqrt{x}}\ dx$$
Here I'd be a bit more ambitious than you.  I know that anything other than simply a variable under a square root is going to cause me trouble so I'm going to try to substitute everything under the outer square root.
Let $u = 1-\sqrt{x}$.  This implies that $$du = \frac{d(1-\sqrt{x})}{dx}\ dx = -\frac{1}{2\sqrt{x}}\ dx = -\frac 1{2(1-u)}\ dx \\ \implies -2(1-u)\ du = dx$$
We also will need to substitute for $x$ so notice that $u=1-\sqrt{x} \implies x=(1-u)^2$.
Now we make this substitution.  Notice how I evaluate the bounds:
$$\begin{align}\int_{x=0}^{x=1} (1-u)^2\sqrt{u}\big(-2(1-u)\ du\big) &= -2\int_{u=1-\sqrt{0}}^{u=1-\sqrt{1}} (1-u)^3\sqrt{u}\ du \\ &= -2\int_{u=1}^{u=0} (1-u)^3u^{1/2}\ du \\ &= 2\int_{u=0}^{u=1} (1-u)^3u^{1/2}\ du\end{align}$$
From here we should just simplify $(1-u)^3u^{1/2}$.  $$(1-u)^3u^{1/2} = (1-3u+3u^2-u^3)u^{1/2} = u^{1/2}-3u^{3/2}+3u^{5/2}-u^{7/2}$$
Now we can use this and the fact that $\int_a^b (\beta f(x) + \gamma g(x))\ dx = \beta\int_a^b f(x)\ dx + \gamma\int_a^b g(x)\ dx$ to get our integral into a form we know how to evaluate.
$$\begin{align}2\int_{u=0}^{u=1} (1-u)^3u^{1/2}\ du &= 2\int_0^1 (u^{1/2}-3u^{3/2}+3u^{5/2}-u^{7/2})\ du \\ &= 2\left(\int_0^1 u^{1/2}\ du -3\int_0^1 u^{3/2}\ du + 3\int_0^1 u^{5/2}\ du -\int_0^1 u^{7/2}\ du\right) \\ &= 2\int_0^1 u^{1/2}\ du -6\int_0^1 u^{3/2}\ du + 6\int_0^1 u^{5/2}\ du - 2\int_0^1 u^{7/2}\ du\end{align}$$
At this point we remember that the antiderivative of $x^k$ is $\frac 1{k+1}x^{k+1}$ and get the answer:
$$2\int_0^1 u^{1/2}\ du -6\int_0^1 u^{3/2}\ du + 6\int_0^1 u^{5/2}\ du - 2\int_0^1 u^{7/2}\ du\ = \left.\big(2\frac 23u^{3/2} -6\frac25u^{5/2} +6\frac27u^{7/2}-2\frac29u^{9/2}\big)\right|_{\ 0}^{\ 1} \\ = \frac 43-\frac{12}5+\frac{12}7-\frac 49$$
A: attempt numero dos, (thanks to @bye_world) (third time maybe charm)
from  $$=2\int_{a}^{b}u^{2}\sqrt{1-u}\:udu$$
$$=2\int_{a}^{b}u^{3}\sqrt{1-u}\:du$$
let $$w=1-u\:$$
 $$u=1-w\:$$
and  $$\frac{dw}{du}=-1$$
 $${-dw}={du}$$
so,  $$=2\int_{b}^{a}({1-w})^{3}\sqrt{w}\:dw$$
 $$=2\int_{b}^{a}(-w^{7/2}+3w^{5/2}-3w^{3/2}+w^{1/2})\:dw$$
Okay, I think I get what you guys are trying very hard to get through to me, $\int{f(x)+g(x)+h(x)}$ can be split up into $\int{f(x)}+\int{g(x)}+\int{h(x)}$ but $\int{f(x)*g(x)}$ does not equal $\int{f(x)}*\int{g(x)}$
Anyways so 
 $$=2[(-\frac{2}{9}(w)^{9/2}+\frac{6}{7}(w)^{7/2}-\frac{6}{5}(w)^{5/2}+\frac{2}{3}(w)^{3/2})]^0_1$$
amirite?
