3

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.

StubbornAtom
  • 17,052
MT_
  • 19,603
  • 9
  • 40
  • 81

4 Answers4

8

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,

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

Barry Cipra
  • 79,832
3

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.

Olivier Oloa
  • 120,989
2

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.

ncmathsadist
  • 49,383
-1

$$\int4x\sqrt{1-x^{4}}dx\\ x^{2}=\sin y\Rightarrow 2xdx=\cos ydy\\ \int4x\sqrt{1-x^{4}}dx=2\int \cos^{2}ydy=\int(1+\cos 2y)dy=y+\frac{\sin 2y}{2}+C\\ \int4x\sqrt{1-x^{4}}dx=\arcsin x^{2}+x^{2}\sqrt{1-x^{4}}+C\\ $$

Bouzari Abdelkader
  • 364