Evaluating Integral

1. Nov 23, 2005

powp

Hello All

I am having problem with solving integrals by subsitution.

I have the following problem. Can anybody help?

(large S) X^2 /(SQRT(1-X)) DX

My first and only thought was to subsitute 1 - x with U but that gave me an answer not ever close.

Does anybody have any tips?

Thanks

P

2. Nov 23, 2005

saltydog

Powp, here it is in LaTex:

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

So let:

$$u=\sqrt{1-x}$$

and work it though completely. That is, then what is $u^2$?

What then is x in terms of u?

What then is dx in terms of u and du?

What is $x^2$ in terms of u?

3. Nov 23, 2005

powp

u^2 = 1 - x
2u * du = -1 * dx
-2u * du = dx

x = 1 - u^2

but where do I go from here.

4. Nov 23, 2005

HallsofIvy

Put them into the integral and DO it!
$$\int \frac{x^2}{\sqrt{1-x}}dx[/itex] If you let $u= \sqrt{1-x}= (1-x)^\frac{1}{2}$ then, yes, $u^2= 1-x$ so 2u du= -dx and x= 1- u2 so x2= (1- u2)2= 1- 2u2+ u4. Your integral becomes [tex]\int \frac{1- 2u^2+ u^4}{u}(-2u du)= -2\int \left(1- 2u^2+ u^4\right)du[/itex]. Actually, your first thought: substituting for 1- x works perfectly well. If u= 1-x then du= -dx. Also, x= 1- u so x2= (1- u)2= 1- 2u+ u2. The integral becomes [tex]-\int \frac{1- 2u+ u^2}{u^\frac{1}{2}}du= -\int \left(u^{-\frac{1}{2}}- 2u^\frac{1}{2}+ u^\frac{3}{2}\right) du$$

Last edited by a moderator: Nov 23, 2005
5. Nov 23, 2005

saltydog

We're makin' progress. Now you need to substitute all these expressions into the original integral:

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

So, substitute each one. I'll do one of them:

$$dx=-2udu$$

There you go. Substitute -2udu for dx in the integral. Well . . . do the same for $x^2$ and $\sqrt{1-x}$ . . . what does the integral then look like in terms of u? It's a lot easier now right? So solve it for u then don't forget to substitute back $\sqrt{1-x}$ in place of u in the final answer.

Edit: Didn't see Hall's reply but you know what to do now.

Last edited: Nov 23, 2005