<s>[INST] Evaluate $\int x \sqrt{1 - x^4} \,\mathrm{d}x$ 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!
 [/INST] Try another substitution: $x^2 = \sin (u)$.
We have $2x dx = \cos (u) du$ so $dx = \frac{\cos (u)}{2x} du$
So now  we have $\frac{1}{2} \displaystyle \int \cos(u) \sqrt{1 - \sin^2 (u)} du$
And I think you can do the rest.
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