$\displaystyle \int x\sqrt{1-x^{4}}dx$

Here's the work I've done, hopefully there's not too much bad math in it...

$\displaystyle x=\sin{\theta}$ so $\displaystyle dx=\cos{\theta}d{\theta}$

$\displaystyle \int \sin{\theta}\sqrt{(1-\sin^2{\theta})^2}\cos{\theta}d{\theta}$

$\displaystyle \int \sin{\theta}\sqrt{(\cos^2{\theta})^2}\cos{\theta}d {\theta}$

$\displaystyle \int \sin{\theta}\cos^2{\theta}\cos{\theta}d{\theta}$

$\displaystyle u=\sin{\theta}$ so $\displaystyle du=\cos{\theta}d{\theta}$

$\displaystyle \int u(1-u^2)du$

$\displaystyle \tfrac{1}{2}u^2 - \tfrac{1}{4}u^4 + C$

...and I stopped right here, seeing that this wouldn't give me the right answer. I'm not ever certain that I can u-sub in this problem like I did. Can someone point me in the right direction please?