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

Question

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

$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.

The first substitution leads to $\pm\int \sin^3\theta\cos^2\theta,d\theta$. — Daniel Fischer, Jul 02 '13 at 19:09
@Dantheman No, $\sqrt{1-x^2}=\cos\theta$ and $dx =\cos\theta ;d\theta$ — Thomas Andrews, Jul 02 '13 at 19:20
I think you are mistaken. I get that x= sintheta, dx=costheta dtheta. Meaning dtheta = dx/costheta. Meaning the cos cancels out and I am let with sin. You forgot that part. — Dantheman, Jul 02 '13 at 19:22
@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 $dx \to \cos\theta,d\theta$. — Daniel Fischer, Jul 02 '13 at 19:25
you should have $ \int \sin(\theta)^3\cos(\theta)^2d\theta. $ — Mhenni Benghorbal, Jul 02 '13 at 19:26
@DanielFischer But dtheta is = dx/costheta. So when I put the dtheta in I have to remove the dx and add a 1/cos which eliminates the cos. — Dantheman, Jul 02 '13 at 19:51
But you don't have a $d\theta$ to start with. You start with a $dx$. And that $dx$ becomes $\cos\theta,d\theta$. — Daniel Fischer, Jul 02 '13 at 19:53
@ThomasAndrews Why does $\sqrt{1-x^2} = \cos\theta$? If $x = \sin\theta$ then $1-x^2 = 1-\sin^2\theta \equiv \cos^2\theta$. Then $\sqrt{1-x^2} = \sqrt{\cos^2\theta} \equiv |\cos\theta|$. For example, what happens when $\theta = \frac{5\pi}{4}$? — Fly by Night, Jul 02 '13 at 19:55
@FlybyNight Yeah, we tend to be a little lazy about such things, not saying so explicitly. Since we substituted freely, we can just say $\theta\in[-\pi/2,\pi/2]$ — Thomas Andrews, Jul 02 '13 at 19:59
It's a pretty common substitution, hence "we" meaning "practically ever other person who has done the same thing, and myself." :) @FlybyNight — Thomas Andrews, Jul 02 '13 at 20:02
No, an infinite domain isn't really permissible, since $x\in[-1,1]$ so we can assume $\theta\in (-\pi/2,\pi/2)$. @FlybyNight — Thomas Andrews, Jul 02 '13 at 20:07

Ron Gordon · Accepted Answer · 2013-07-02T20:31:18.897

3

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

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

edited Jul 02 '13 at 20:31

answered Jul 02 '13 at 20:07

Ron Gordon

138,521

You seem to be apprehending identities out of thin air. Where did The first two lines come from? I don't follow at all. – Dantheman Jul 02 '13 at 20:13
@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. – Ron Gordon Jul 02 '13 at 20:14
I don't see the identity. Why is sin cubed time cosin squared equal to integral sin cubed - integral sin to the fifth? – Dantheman Jul 02 '13 at 20:19
@Dantheman What Ron did was: $$ \int \sin^3\theta\cos^2\theta d\theta = \int \sin^3\theta(1 - \sin^2\theta) d\theta = \int \sin^3\theta - \sin^5\theta d\theta $$ – Adriano Jul 02 '13 at 20:28
@Dantheman: $$\sin^3{\theta} \cos^2{\theta} = \sin^3{\theta} (1-\sin^2{\theta})$$ – Ron Gordon Jul 02 '13 at 20:28
Is that easier than doing the standard (1-cos^2x)^k cos^nx sinx dx? integral – Dantheman Jul 02 '13 at 20:35
I've cleared up the discussion about notation. – Michael Greinecker Jul 02 '13 at 22:43

DonAntonio · Answer 2 · 2013-07-02T22:48:48.307

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:

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

score 1 · Answer 3 · answered Jul 02 '13 at 20:03

1

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 $x$...

answered Jul 02 '13 at 20:03

DJohnM

3,580

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 – Dantheman Jul 02 '13 at 20:18
Sure, but don't forget, part of learning an integration technique is learning when, and when not, to use it :) – DJohnM Jul 02 '13 at 20:29
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. – Triatticus Jul 02 '13 at 21:00
There are too many to remember for a summer class. – Dantheman Jul 02 '13 at 21:01

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

3 Answers3