1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Evaluating integral

  1. Sep 24, 2013 #1
    1. The problem statement, all variables and given/known data
    If ##y(x-y)^2=x##, then find
    $$\int \frac{dx}{x-3y}$$


    2. Relevant equations



    3. The attempt at a solution
    I tried expanding the given equation. That gives me a cubic in y. I found no way to solve the cubic equation. I plugged the equation in Wolfram Alpha to see if it gives any nice solution but nope, no luck there too. Now I think that this is some sort of trick question but I don't really know where to begin.

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Sep 24, 2013 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    What about changing to the variable y/x=t? Just a guess...

    ehild
     
  4. Sep 24, 2013 #3
    I should have really attempted the problem properly before posting it here. I just woke up and gave it a try. I substituted x-y=t. The integral changes to dx/(3t-2x) and the equation changes to ##(x-t)t^2=x##. I differentiated the equation wrt x and after some simplification, I got
    $$\frac{dx}{3t-2x}=\frac{dt}{t^2-1}$$

    Now its a very easy integral to solve. Thanks for the effort ehild. :smile:
     
  5. Sep 24, 2013 #4
    I think that's wrong. Sides, once you got t, then what? How about checking it with the integral:

    [tex]\int_1^2 \frac{1}{x-3y}dx,\quad y(x-y)^2=x[/tex]

    While we're at it, what is the geometric interpretation of that integral anyway?
     
  6. Sep 24, 2013 #5
    I don't see what's wrong here. I differentiated both the sides of the equation wrt x. Did I do some mistake while simplifying the expression?
     
  7. Sep 25, 2013 #6
    Just realised that the final equation after differentiating I get is wrong. It is actually
    $$\frac{dx}{3t-2x}=\frac{t \, dt}{t^2-1}$$
     
  8. Sep 25, 2013 #7
    To check it, I solved the definite integral above and obtained:

    [tex]\int_1^2 \frac{1}{x-3y} dx=\int_a^b \frac{t}{t^2-1}dt[/tex]

    where a=1-lowroot and b=2-highroot with the roots being the real roots of the equations [itex]y(1-y)^2=1[/itex] and [itex]y(2-y)^2=2[/itex]

    which agrees numerically with direct substitution of y(x) from the solution of [itex]y(x-y)^2=x[/itex].

    Always a good idea to check them numerically if we're trying to intercept an earth-bound meteor.
     
  9. Sep 25, 2013 #8
    Given ::##y\cdot(x-y)^2 = x##


    Now Diff. both side w. r . to ##x## , We Get


    ##\displaystyle y\cdot 2(x-y)\left\{1-\frac{dy}{dx}\right\}+(x-y)^2 \cdot \frac{dy}{dx} = 1##


    ##\displaystyle 2y\cdot(x-y)\left\{1-\frac{dy}{dx}\right\}+\frac{dy}{dx}\cdot (x-y)^2-(x-y)^2 = 1-(x-y)^2##


    ##\displaystyle \left\{1-\frac{dy}{dx}\right\}\left(2xy-y^2-x^2+2xy-y^2\right) = 1-(x-y)^2##


    ##\displaystyle \left\{1-\frac{dy}{dx}\right\}\left(-x^2+4xy-3y^2\right) = 1-(x-y)^2##


    ##\displaystyle \left\{\frac{dy}{dx}-1 \right\}\left(x^2-4xy+3y^2\right) = 1 - (x-y)^2##


    ##\displaystyle \left\{\frac{dy}{dx}-1 \right\}.\left(x-y\right).\left(x-3y\right) = 1-(x-y)^2##


    ##\displaystyle \left(dy - dx\right).(x-y).(x-3y) = \left\{1-(x-y)^2\right\}dx##


    ##\displaystyle \frac{dx}{(x-3y)} = \frac{\left(dy - dx\right)\cdot(x-y)}{1-(x-y)^2}##


    Now Integrate both side , We Get


    ##\displaystyle \int\frac{dx}{(x-3y)} = \int \frac{\left(dx - dy\right)\cdot(x-y)}{(x-y)^2 -1}##


    Let ##(x-y) = t\Leftrightarrow d(x-y) = dt\Leftrightarrow \left(dx - dy\right) = dt##


    So ##\displaystyle \int\frac{dx}{(x-3y)} = \int \frac{t}{t^2-1}dt##


    So again Let ##\displaystyle t^2 - 1 = u\Leftrightarrow tdt = \frac{1}{2}du##


    So ##\displaystyle \int\frac{dx}{(x-3y)} = \frac{1}{2}\cdot \int\frac{1}{u}du = \frac{1}{2}\cdot \ln \left|u\right|+C##


    So ##\displaystyle \int\frac{dx}{(x-3y)} = \frac{1}{2}\cdot \ln \left|(x-y)^2 - 1\right|+C##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Evaluating integral
  1. Evaluate this integral (Replies: 2)

  2. Evaluate the Integral (Replies: 5)

  3. Evaluate this integral (Replies: 4)

  4. Evaluate the integral (Replies: 7)

  5. Integral evaluation (Replies: 5)

Loading...