How to Solve Integrals by Substitution: Tips and Tricks

[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

 

[saltydog]

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

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?

 

[powp]

Thanks for the reply

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

x = 1 - u^2

but where do I go from here.

 

[HallsofIvy]

Put them into the integral and DO it!
Your original integral was
\int \frac{x^2}{\sqrt{1-x}}dx[/itex]<br /> 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<br /> \int \frac{1- 2u^2+ u^4}{u}(-2u du)= -2\int \left(1- 2u^2+ u^4\right)du[/itex].&lt;br /&gt; Actually, your first thought: substituting for 1- x works perfectly well.&lt;br /&gt; If u= 1-x then du= -dx. Also, x= 1- u so x&lt;sup&gt;2&lt;/sup&gt;= (1- u)&lt;sup&gt;2&lt;/sup&gt;= 1- 2u+ u&lt;sup&gt;2&lt;/sup&gt;. The integral becomes &lt;br /&gt; -\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

 

[saltydog]

Thanks for the reply
u^2 = 1 - x
2u * du = -1 * dx
-2u * du = dx
x = 1 - u^2
but where do I go from here.

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.