# Definite integral substitution???

• May 18th 2008, 07:34 PM
rexexdesign
Definite integral substitution???
Hi all,

I am struggling to find a way to substitution for integral of x^2/sqr(1-x)dx.
If it was x^3 in the square root I would know how to do it, but because numerator is lower power than denominator

Thanks
alex
• May 18th 2008, 07:42 PM
Mathstud28
Quote:

Originally Posted by rexexdesign
Hi all,

I am struggling to find a way to substitution for integral of x^2/sqr(1-x)dx.
If it was x^3 in the square root I would know how to do it, but because numerator is lower power than denominator

Thanks
alex

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

let $\displaystyle u=\sqrt{1-x}\Rightarrow{x=1-u^2}$

So $\displaystyle dx=-2udu$

and $\displaystyle x^2=(1-u^2)^2$

So now we have

$\displaystyle \int\frac{-2u(1-u^2)^2}{u}du$

I think you can take it from there
• May 18th 2008, 08:23 PM
rexexdesign
You are right. My teacher did not teach us that before. She always said find the outside and inside function and make substitutions, like $\displaystyle \sqrt{x-1}\$ would be $\displaystyle f(x)=\sqrt{x}\$ and $\displaystyle g(x)=x-1$

thanks for the tip.
• May 18th 2008, 08:33 PM
Mathstud28
Quote:

Originally Posted by rexexdesign
You are right. My teacher did not teach us that before. She always said find the outside and inside function and make substitutions, like $\displaystyle \sqrt{x-1}\$ would be $\displaystyle f(x)=\sqrt{x}\$ and $\displaystyle g(x)=x-1$

thanks for the tip.

It is an art form though. As I am sure most will attest on this website it takes a while to see whether you should for this example

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

or Let $\displaystyle u=1-x$