6
$\begingroup$

$$\int x^3 \sqrt{1-x^2} dx$$

$x = \sin \theta $

$dx = \cos \theta d \theta$

$$\int \sin^3 \theta d \theta$$

$$\int (1 - \cos^2 \theta) \sin \theta d \theta$$

$u = \cos \theta$

$du = -\sin\theta d \theta$

$$-\int u^2 du$$

$$\frac{-u^3}{3} $$

$$\frac{\cos^3 \theta}{3}$$

With the triangle trick I get:

$$\frac{-\sqrt{1-x^2}^3}{3}$$

This is wrong but I am not sure where I went wrong.

$\endgroup$
  • 5
    $\begingroup$ The first substitution leads to $\pm\int \sin^3\theta\cos^2\theta\,d\theta$. $\endgroup$ – Daniel Fischer Jul 2 '13 at 19:09
  • $\begingroup$ Shouldn't that be just cos? $\endgroup$ – Dantheman Jul 2 '13 at 19:16
  • 1
    $\begingroup$ @Dantheman You have $x^3 \cdot \sqrt{1-x^2} \cdot dx$. you replace each factor with the corresponding result of the substitution $x \to \sin\theta$, and $dx \to \cos\theta\,d\theta$. $\endgroup$ – Daniel Fischer Jul 2 '13 at 19:25
  • 1
    $\begingroup$ But you don't have a $d\theta$ to start with. You start with a $dx$. And that $dx$ becomes $\cos\theta\,d\theta$. $\endgroup$ – Daniel Fischer Jul 2 '13 at 19:53
  • 1
    $\begingroup$ It's a pretty common substitution, hence "we" meaning "practically ever other person who has done the same thing, and myself." :) @FlybyNight $\endgroup$ – Thomas Andrews Jul 2 '13 at 20:02
3
$\begingroup$

Let $x=\sin{\theta}$, then $dx = \cos{\theta} \, d\theta$; the integral becomes

$$\int d\theta \, \sin^3{\theta} \, \cos^2{\theta} = \int d\theta \, \sin^3{\theta} -\int d\theta \, \sin^5{\theta} $$

$$\int d\theta \, \sin^3{\theta} = \int d\theta \, \sin{\theta} (1-\cos^2{\theta}) = -\int d(\cos{\theta}) (1-\cos^2{\theta})= -\cos{\theta} + \frac13 \cos^3{\theta}+C$$

Similarly

$$\int d\theta \, \sin^5{\theta} = -\int d(\cos{\theta}) (1-\cos^2{\theta})^2 = -\cos{\theta} + \frac{2}{3} \cos^3{\theta}-\frac15 \cos^5{\theta}+C'$$

Subtracting the two, I get

$$\int d\theta \, \sin^3{\theta} \, \cos^2{\theta} = -\frac13 \cos^3{\theta}+\frac15 \cos^5{\theta}+C$$

Then use $x=\sin{\theta}$ and get

$$\int dx \, x^3 \, \sqrt{1-x^2} = \frac{1}{15} (3 x^4-x^2-2) \sqrt{1-x^2}+C$$

EDIT

I see that the answer can be simplified further to

$$-\frac{1}{15} (1-x^2)^{3/2} (3 x^2+2) + C$$

$\endgroup$
  • $\begingroup$ You seem to be apprehending identities out of thin air. Where did The first two lines come from? I don't follow at all. $\endgroup$ – Dantheman Jul 2 '13 at 20:13
  • $\begingroup$ @Dantheman: the only identities I use is $\cos^2{\theta} = 1-\sin^2{\theta}$ and d(\cos{\theta}) = -\sin{\theta} d\theta$. The rest is algebra. $\endgroup$ – Ron Gordon Jul 2 '13 at 20:14
  • $\begingroup$ I don't see the identity. Why is sin cubed time cosin squared equal to integral sin cubed - integral sin to the fifth? $\endgroup$ – Dantheman Jul 2 '13 at 20:19
  • $\begingroup$ @Dantheman What Ron did was: $$ \int \sin^3\theta\cos^2\theta d\theta = \int \sin^3\theta(1 - \sin^2\theta) d\theta = \int \sin^3\theta - \sin^5\theta d\theta $$ $\endgroup$ – Adriano Jul 2 '13 at 20:28
  • $\begingroup$ @Dantheman: $$\sin^3{\theta} \cos^2{\theta} = \sin^3{\theta} (1-\sin^2{\theta})$$ $\endgroup$ – Ron Gordon Jul 2 '13 at 20:28
2
$\begingroup$

What about by integration by parts? It looks pretty simple:

$$u=x^2\;,\;\;u'=2x\\v'=x\sqrt{1-x^2}\;,\;\;v=-\frac13(1-x^2)^{3/2}$$

and thus

$$\int x^2\cdot x\sqrt{1-x^2}\,dx=-\frac13x^2(1-x^2)^{3/2}+\frac23\int x(1-x^2)^{3/2}dx=$$

$$-\frac13x^2(1-x^2)^{3/2}-\frac2{15}(1-x^2)^{5/2}+C$$

Note: We used above the following:

$$\int x(1-x^2)^k\,dx=-\frac12\int (-2x\,dx)(1-x^2)^k=-\frac12\frac{(1-x^2)^{k+1}}{k+1}\;\ldots$$

$\endgroup$
1
$\begingroup$

You can avoid all the trig by making a much simpler substitution:$$u^2=1-x^2$$so: $$x^2=1-u^2$$ $$ u= \sqrt{1-x^2}$$ $$2u du=-2xdx$$

Rewriting the integral, factoring out one $x$:$$\int x^3 \sqrt{1-x^2} dx=\int x^2 \sqrt{1-x^2} xdx=-\int (1-u^2) u^2 du$$Multiply out the integrand, integrate with the power formula term by term and substitute back for $x$...

$\endgroup$
  • $\begingroup$ I would prefer to use trig substitution since I need to learn it. If I wanted to do it the easy way I would have put it into wolfram alpha :P $\endgroup$ – Dantheman Jul 2 '13 at 20:18
  • $\begingroup$ Sure, but don't forget, part of learning an integration technique is learning when, and when not, to use it :) $\endgroup$ – DJohnM Jul 2 '13 at 20:29
  • $\begingroup$ I tell everyone when this chapter comes to start learning their trig identities they forgot back in 10th grade, it doesn't hurt to prove some of them to oneself to understand where they all come from. I remember one time it saved me on a quantum mechanics assignment to remember the cotangent and tangent half angle identities. $\endgroup$ – Triatticus Jul 2 '13 at 21:00
  • $\begingroup$ There are too many to remember for a summer class. $\endgroup$ – Dantheman Jul 2 '13 at 21:01

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.