5
$\begingroup$

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!

$\endgroup$
  • 4
    $\begingroup$ What about $x^2=\sin(y)$ ? $\endgroup$ – Claude Leibovici Apr 17 '14 at 14:26
  • $\begingroup$ Well...that simplifies everything a whole lot - doesn't it? Thanks! $\endgroup$ – Jeel Shah Apr 17 '14 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. $u=\cos\theta.$ $\endgroup$ – Américo Tavares Apr 17 '14 at 14:44
  • 1
    $\begingroup$ @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 '14 at 14:51
4
$\begingroup$

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.

$\endgroup$
11
$\begingroup$

HINT :

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.

ADDENDUM :

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.