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?


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!

  • 4
    $\begingroup$ What about $x^2=\sin(y)$ ? $\endgroup$ Apr 17, 2014 at 14:26
  • $\begingroup$ Well...that simplifies everything a whole lot - doesn't it? Thanks! $\endgroup$
    – Jeel Shah
    Apr 17, 2014 at 14:28
  • 2
    $\begingroup$ You coud try first the substitution $u=x^2$. Your integral becomes $\frac{1}{2}\int\sqrt{1-u^2}du$. Then you could use e.g. $u=\cos\theta.$ $\endgroup$ Apr 17, 2014 at 14:44
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    $\begingroup$ @AméricoTavares That is a good idea, since that is essentially where the motivation for using $\cos (u) = x^2$ comes from (it's important to learn the motivation IMO) $\endgroup$
    – MCT
    Apr 17, 2014 at 14:51

2 Answers 2



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.


I have just found that the fastest way is letting $u=x^2$.

  • 1
    $\begingroup$ @AméricoTavares Thank you Sir. I just think too complicated for the first method. $\endgroup$
    – Tunk-Fey
    Apr 17, 2014 at 14:47
  • $\begingroup$ that addendum is really elegant $\endgroup$ Apr 17, 2014 at 15:44
  • $\begingroup$ Nice. +1 @Tunk-Fey $\endgroup$ Apr 17, 2014 at 16:50
  • $\begingroup$ @Integrals Thank you Jeff. Also thank you frogeyedpeas. :) $\endgroup$
    – Tunk-Fey
    Apr 17, 2014 at 16:55

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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