• Support PF! Buy your school textbooks, materials and every day products Here!

Math experts please come

  • #1
Math experts please come!!

b]1. Homework Statement [/b]

Calculus, integration

Homework Equations


http://img341.imageshack.us/img341/6091/questiontc3.jpg [Broken]


The Attempt at a Solution



used method:changing variable, by part, trigo etc but no avail.:confused: Please help me solve it, thanks.
 
Last edited by a moderator:

Answers and Replies

  • #2
150
0
SD, Your kidding, right? Didn't your high school math teacher make you do a "time out" whenever misused the square root that way?
 
Last edited:
  • #3
150
0
Rewrite the argument x^2-x by "completing the square" . "u" substitution follows.
 
  • #4
Last edited by a moderator:
  • #5
VietDao29
Homework Helper
1,423
1
When you meet the problem asking you to integrate something like:
[tex]\int \frac{dx}{(ax + b) ^ {m} \sqrt{c x ^ 2 + dx + e}}[/tex], what you should do is to let:
[tex]ax + b = \frac{1}{t}[/tex]
--------------------
So, here, we go:
[tex]\int \frac{dx}{(2x + 1) \sqrt{x ^ 2 - x}}[/tex]
For [tex]\sqrt{x ^ 2 - x}[/tex] to be valid, x can only be in the interval [tex]] - \infty ; 0 ] \unity [1 ; + \infty [ [/tex]
Let
[tex]2x + 1 = \frac{1}{t} \quad \mbox{or} \quad x = \frac{1}{2} \left( \frac{1}{t} - 1 \right)[/tex]
[tex]\Rightarrow 2dx = -\frac{dt}{t ^ 2}[/tex]
The integral becomes:
[tex]- \int \frac{\frac{dt}{2 t ^ 2}}{\frac{1}{t} \sqrt{ \left[ \frac{1}{2} \left( \frac{1}{t} - 1 \right) \right] ^ 2 - \frac{1}{2} \left( \frac{1}{t} - 1 \right)}}[/tex]

[tex]= - \int \frac{dt}{2t\sqrt{ \left[ \frac{1}{4} \left( \frac{1}{t ^ 2} - \frac{2}{t} + 1 \right) \right] - \left( \frac{1}{2t} - \frac{1}{2} \right)}}[/tex]

[tex]= - \int \frac{dt}{2t\sqrt{\frac{1}{4 t ^ 2} - \frac{1}{t} + \frac{3}{4}}}[/tex]

[tex]= - \int \frac{dt}{t \sqrt{\frac{1}{t ^ 2} - \frac{4}{t} + 3}}[/tex]

[tex]= \pm \int \frac{dt}{\sqrt{1 - 4t + 3 t ^ 2}}[/tex]
Can you go from here? :)
 
Last edited:
  • #6
150
0
Forget what I had written here
 
Last edited:
  • #7
I vote for Man Monkey's approach.
Won't work because the integrand has dx/(2x-1) and not (2x-1)dx.

The best approach here is partial fraction decomp.

Let 1=A(2x-1) + B(x^2-x)^1/2

Set x=1 , A=1
Set x=1/2 , B=2i

Then integrate the new functions 1/(x^2-x)^1/2 and 2i/(2x-1)

EDIT: The function 1/(x^2-x)^1/2 can be broken down further into |x^1/2|(x-1)^1/2 or into ((x-x^1/2)(x+x^1/2))^1/2, though the first of these two should look familiar.
 
Last edited:
  • #8
30
0
(x^2-x)^(1/2) = x - x^(1/2) is not true. You can check by multiplying the RHS with itself.
 
  • #9
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
131
What they're equivalent, just trying to help? Thought it might be easier if he got rid of the sqrt.:smile:
[tex]\sqrt{4^{2}-4}=\sqrt{16-4}=\sqrt{12}=2=4-\sqrt{4}[/tex]
Fascinating..
 
  • #10
[tex]\sqrt{4^{2}-4}=\sqrt{16-4}=\sqrt{12}=2=4-\sqrt{4}[/tex]
Fascinating..
OK my bad. I was thinking of [tex]\sqrt{x^2} - \sqrt{x}[/tex]

In my defence I had just finished a long and arduous day at work and wasn't really thinking:eek: sorry.

I deleted it, still the shame lives on.... I'll go bury myself under a pile of soil and pretend I'm dead :smile:

For my next trick I'll prove Fermat's last theorem in a new and exciting way whilst drunk :/
 
Last edited:
  • #11
150
0
going with my first reply

Opps, I got fooled. Monkey Man's answer seemed so slick. I now vote for my own approach. It boils down to a constant times integral of
1/(u sqrt(u^2 -a^2))


"complete the square."

I don't trust partial fractions since we don't have a rational function.
 
  • #12
I don't trust partial fractions since we don't have a rational function.
Your kidding right? So 1/((2x-1)(x^2-x)^1/2) is not a rational function?
 
  • #13
150
0
Nope, It's not. Rational functions are ratios of polynomials.
 
  • #14
So 1/x is no longer rational?
 
  • #15
150
0
is too

(no square root in that one)
 
  • #16
what does the square root matter?
 
  • #17
In that case, u=2x-1

1/u(((u+1)2)^2-(u+1)/2)^1/2
 
  • #18
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
  • #19
Thanks vietdao29. Ya i can go from there...I am a form 4 student studying integration on my own. Hope these can help me in form 5 and form 6.
 
  • #20
150
0
There is an easier way. I had the quickest way expect that I got diss'ed over partial fractions. You shouldn't have to rely on megaformulas.
 
  • #21
682
1
[tex]u=\sqrt{x^2-x}[/tex]
[tex]du=\frac{(2x-1)dx}{2\sqrt{x^2-x}}[/tex]
[tex]\int{\frac{dx}{(2x-1)\sqrt{x^2-x}}}=\int{\frac{2du}{(2x-1)^2}}=\int{\frac{2du}{4u^2+1}}[/tex]
 
Last edited:
  • #22
150
0
I agree except for the last step.
 
  • #23
150
0
In case no one was paying attention, see my attachment for the quick way.
 

Attachments

Last edited:
  • #24
Gib Z
Homework Helper
3,346
5
[tex]u=\sqrt{x^2-x}[/tex]
[tex]du=\frac{(2x-1)dx}{2\sqrt{x^2-x}}[/tex]
[tex]\int{\frac{dx}{(2x-1)\sqrt{x^2-x}}}=\int{\frac{2du}{(2x-1)^2}}=\int{\frac{2du}{4u+1}}[/tex]
[tex]\int{\frac{2du}{(2x-1)^2}}=\int{\frac{2du}{(2u)^2+1}}[/tex] Very easy now, form of arctan integral.
 
  • #25
150
0
I don't agree. Forgetting what "u" is and applying bad algebra. Is anyone reading my attachment?
 

Related Threads on Math experts please come

  • Last Post
Replies
7
Views
9K
Replies
1
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
4
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
10
Views
931
  • Last Post
Replies
1
Views
1K
Replies
15
Views
2K
Replies
23
Views
2K
Top