# 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.

## 3 Answers

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,

$$\int4x\sqrt{1-x^4}\,dx=\int2\cos^2\theta\,d\theta=\int(1+\cos2\theta)\,d\theta\\ =\theta+{1\over2}\sin2\theta+C\\ =\theta+\sin\theta\cos\theta+C\\=\arcsin(x^2)+x^2\sqrt{1-x^4}+C$$

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.

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.