TITLE: Evaluate $\int x \sqrt{1 - x^4} \,\mathrm{d}x$
QUESTION [5 upvotes]: I have the following question
$$\int x \sqrt{1 - x^4} \,\mathrm{d}x$$
I know we have to use trig. substitution for this and therefore, I did the following by letting $x = \sin \theta$ and $dx = \cos \theta \,\mathrm{d}\theta$
\begin{align}
&\int x \sqrt{1-x^4} \,\mathrm{d}x \\
&=\int \sin \theta \cos \theta\sqrt{1 - (\sin \theta)^4} \, \mathrm{d}\theta \\
&=\int \sin \theta \cos \theta \sqrt{(1-\sin^2 \theta)(1+\sin^2\theta)} \,\mathrm{d}\theta \\
&=\int \sin \theta \cos \theta \sqrt{(\cos^2 \theta)(1+\sin^2\theta)} \,\mathrm{d}\theta \\
\end{align}
Now, I'm confused. How do I proceed?
Thanks!
EDIT: Taking from the answer, I have a (nearly) full solution below for future users.
Instead of letting $x = \sin \theta$. We'll let $x^2 = \sin \theta$ and this will greatly simplify everything. Since, $\mathrm{d}x = \frac{\cos \theta}{2x} \,\mathrm{d\theta}$
\begin{align}
&\int x \sqrt{1-x^4} \quad \mathrm{d}x \\
&=\frac{1}{2}\int \cos \theta\sqrt{1 - (\sin x)^2} \quad \mathrm{d}\theta \\
&= \frac{1}{2} \int \cos \theta \cos \theta \quad \mathrm{d}\theta \\
&= \frac{1}{4} \int 1 + \cos 2\theta \quad \mathrm{d}\theta \\
&= \frac{1}{4} \left(\theta + \frac{\sin2\theta}{2} \right) \\
&= \frac{\theta}{4} + \frac{\sin \theta \cos \theta}{4} \\
\end{align}
After this, you only have to put $\sin \theta$ back in terms of $x$ and you're done!

REPLY [10 votes]: HINT :
Let $u^2=1-x^4\;\Rightarrow\; x^2=\sqrt{1-u^2}\;\Rightarrow\; x\ dx=-\dfrac{u\ du}{2\sqrt{1-u^2}}$, then rewrite
$$
\begin{align}
\int x\sqrt{1-x^4}\ dx&=-\frac12\int\dfrac{u^2}{\sqrt{1-u^2}}\ du\\
&=\frac12\int\dfrac{1-u^2-1}{\sqrt{1-u^2}}\ du\\
&=\frac12\left(\int\dfrac{1-u^2}{\sqrt{1-u^2}}\ du-\int\dfrac{1}{\sqrt{1-u^2}}\ du\right).
\end{align}
$$
The left part integral can be solved by using IBP and the right part integral can be solved by  using trigonometry substitution.
ADDENDUM :
I have just found that the fastest way is letting $u=x^2$.