Indefinite Integration by u-sub/trig sub

[wick3dgt]

Homework Statement

Integrate (x^3)sqrt(1-x^2)

Homework Equations

The Attempt at a Solution

I used trig. substitution along with u substitution and came up with (x^4)/4 +C which I know is wrong. My professor gave the answer -(((3x^2)+2)((1-x^2)^(3/2)))/15 . Please help!

 

[wick3dgt]

Homework Statement

Integrate (x^3)sqrt(1-x^2)

Homework Equations

The Attempt at a Solution

I used trig. substitution along with u substitution and came up with (x^4)/4 +C which I know is wrong. My professor gave the answer -(((3x^2)+2)((1-x^2)^(3/2)))/15 . Please help!

 

[quasar987]

Is there a special restriction in the question that says you can't use integration by parts?

 

[Dick]

The substitution you want is u=1-x^2. One x from the x^3 will then be used to create the du leaving x^2 but then you'll remember that x^2=1-u. Try it, you'll like it!

 

[Gib Z]

Which substitution? x=sin theta should work fine.

 

[Gib Z]

Ok I worked through it for you.
$$\int \frac{x^3}{\sqrt{1-x^2}} dx$$
x=sin theta
dx=cos theta d(theta)
$$\int \frac{x^3}{\sqrt{1-x^2}} dx = \int \sin^3 \theta d\theta$$
$$=\int \sin^2 \theta \cdot \sin \theta d\theta = \int (1-\cos^2 \theta)\sin \theta d\theta = \int \sin \theta d\theta - \int \cos^2 \theta \sin \theta d\theta$$

First integral is just -cos theta, 2nd integral is simple with u=cos theta. Easy as pi.

 

[TheoMcCloskey]

Yep - I get the Prof's result too.

First substitution: $$u= x^2$$

Second substition: let $$v^2 = 1-u$$

You should wind up with an integral of:

I = $$\int (v^4-v^2) dv$$

 

[neutrino]

Ok I worked through it for you.
$$\int \frac{x^3}{\sqrt{1-x^2}} dx$$
x=sin theta
dx=cos theta d(theta)
$$\int \frac{x^3}{\sqrt{1-x^2}} dx = \int \sin^3 \theta d\theta$$
$$=\int \sin^2 \theta \cdot \sin \theta d\theta = \int (1-\cos^2 \theta)\sin \theta d\theta = \int \sin \theta d\theta - \int \cos^2 \theta \sin \theta d\theta$$

First integral is just -cos theta, 2nd integral is simple with u=cos theta. Easy as pi.

Only if it were that simple. The radical is in the numerator.

 

[Gib Z]

I can't read. Kill me. I'll get another solution for you tomorrow then, for the right problem :P

 

[Gib Z]

Ok actually no problem either way.
Same substitution, x=sin theta
dx= cos theta d(theta)

$$\int x^3 \sqrt{1-x^2} dx = \int \sin^3 \theta \cos^2 \theta d\theta = \int \sin \theta (1-\cos^2 \theta)\cos^2 \theta d\theta = \int (\cos^2 \theta - \cos^4 \theta) \sin \theta d\theta$$ Which is very simple with the easy substitution, u=cos x.

EDIT: Rewritten in the form $$\int (\cos^4 \theta - \cos^2 \theta) (-\sin \theta) d\theta$$ it becomes what TheoMcCloskey said it would be.

 

[HallsofIvy]

But as Dick and ThomasMcCloskey (indirectly) said, The substitution u= 1-x² is simpler.