Evaluating the indefinite integral $ \int 4x \sqrt{1 - x^4} \,dx$ I need help evaluating $$\int 4x \sqrt{1 - x^4} dx$$
What I have tried so far: Rewriting the integral as $$\int \frac{4x}{\sqrt{1 - x^4}} (1 - x^4) dx$$
$$\int \frac{4x}{\sqrt{1 - x^4}}dx - \int \frac{4x^5}{\sqrt{1 - x^4}} dx$$
The first integral I can evaluate using substituting $t = x^2, dt = 2xdx$
$$\int \frac{2}{\sqrt{1 - t^2}} dt = 2 \sin^{-1} t$$
I tried the same substitution on the second integral:
$$\int \frac{2t^2}{\sqrt{1 - t^2}}dt $$
But now I am stuck. Am I going in the right direction?
edit: trying out integration by parts. That just struck me.
 A: I recommend starting from scratch, with the substitution $x^2=\sin\theta$, so that $2x\,dx=\cos\theta \,d\theta$ and $\sqrt{1-x^4}=\sqrt{1-\sin^2\theta}=\cos\theta$.  Thus, using a couple of other standard trig identities,
$$\int4x\sqrt{1-x^4}\,dx=\int2\cos^2\theta\,d\theta=\int(1+\cos2\theta)\,d\theta\\
=\theta+{1\over2}\sin2\theta+C\\ =\theta+\sin\theta\cos\theta+C\\=\arcsin(x^2)+x^2\sqrt{1-x^4}+C$$
A: You may write
$$
\begin{align}
\int 4x \sqrt{1 - x^4} dx & =2\int 2x \sqrt{1 - (x^2)^2} dx \\\\
& =2\int \sqrt{1 - u^2} du \\\\
& =2\int \cos t \:\sqrt{1 - \sin^2 t} \: dt \\\\
& =2\int \cos^2 t \: dt \\\\
& =t+\frac12 \sin (2t)+C\\\\
& =t+\sin t \cos t+C\\\\
& =\arcsin (x^2)+x^2\:\sqrt{1 - x^4}+C,\\\\
\end{align}
$$
on an appropriate interval.
A: Yes.  You can try one of two things. Integrate by parts with
$$u = t$$ and 
$$dv = {2t\,dt\over \sqrt{1 - t^2}}$$
or a trig sub $t = \sin(\theta)$.  The or is inclusive or here.
A: $$\int4x\sqrt{1-x^{4}}dx\\
x^{2}=\sin y\Rightarrow 2xdx=\cos ydy\\
\int4x\sqrt{1-x^{4}}dx=2\int \cos^{2}ydy=\int(1+\cos 2y)dy=y+\frac{\sin 2y}{2}+C\\
\int4x\sqrt{1-x^{4}}dx=\arcsin x^{2}+x^{2}\sqrt{1-x^{4}}+C\\
$$
