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Integral - best way to solve

  1. Aug 14, 2014 #1

    etf

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    Here is my integral:
    $$\\\int \frac{y}{((x-1)^{2}+y^{2})^{2}}dy$$
    I solved it using substitution $$y=(x-1)\tan{\varphi }$$
    Final result is $$\frac{1}{2(x-1)^{2}}\sin^{2}{(arctan(\frac{y}{x-1}))}$$
    It was complicated to solve it and I used few trig. identities. My question is, is it possible to solve it using another method (faster)?
     
  2. jcsd
  3. Aug 14, 2014 #2

    LCKurtz

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    I would just call ##(x-1)^2=a^2## so you have ##\int \frac y {(a^2+y^2)^2}~dy## and let ##u = a^2 +y^2##, ##du = 2y~dy##. Your answer doesn't look even close to correct.
     
  4. Aug 14, 2014 #3

    mfb

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    Well, $$\sin^2 (\arctan (z)) = \frac{z^2}{z^2+1}$$The answer is not correct, but it is not far away from the right answer (just expressed in a very complicated way).
     
  5. Aug 14, 2014 #4

    etf

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    I found derivative of my solution (using Matlab) and I got my original function :smile: Are you sure it's not correct?
     
  6. Aug 14, 2014 #5

    LCKurtz

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    Well, since you didn't show your work I didn't spend any time trying to verify it. Your method is certainly not the preferred method for this problem. Try the substitution I suggested and you will see.
     
  7. Aug 14, 2014 #6

    etf

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    I will post complete solution.
     
  8. Aug 14, 2014 #7

    mfb

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    I don't get that. Note that I replaced x-1 by x as it always appears in this combination.
     
  9. Aug 14, 2014 #8

    etf

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    $$\int \frac{y}{((x-1)^{2}+y^{2})^{2}}dy=\left \{ y=(x-1)tan\varphi ,dy=\frac{x-1}{\cos^{2}{\varphi }}d\varphi \right \}=$$$$\\\int \frac{(x-1)tan\varphi }{((x-1)^{2}+(x-1)^{2}tan^{2}\varphi )^{2}}\frac{x-1}{\cos^{2}{\varphi }}d\varphi =$$$$\\\int \frac{(x-1)tan\varphi }{(x-1)^{4}(1+tan^{2}\varphi )^{2}}\frac{x-1}{\cos^{2}{\varphi }}d\varphi =$$$$\\\frac{1}{(x-1)^{2}}\int \frac{tan\varphi }{(sec^{2}\varphi )^{2}}\frac{1}{\cos^{2}{\varphi }}d\varphi =$$ $$\\=\frac{1}{(x-1)^{2}}\int \frac{\frac{\sin{\varphi }}{\cos{\varphi }}}{\frac{1}{\cos^{4}{\varphi }}\cos^{2}{\varphi }}d\varphi=\frac{1}{(x-1)^{2}}\int \sin{\varphi }cos{\varphi}d\varphi =\frac{1}{2(x-1)^{2}}\int \sin{2\varphi }d\varphi =\frac{1}{2(x-1)^{2}}\sin^{2}{\varphi }$$
     
    Last edited: Aug 14, 2014
  10. Aug 14, 2014 #9

    etf

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    You forgot to raise sin at 2 ? If you put d/dy (1/(2x^2) * (sin(arctan(y/x)))^2) you will get correct result...
    Here is my result from Matlab:
    >> syms x y
    >> f=(1/(2*(x-1)^2))*(sin(atan(y/(x-1))))^2;
    >> simplify(diff(f,y)-y/((x-1)^2+y^2)^2)

    ans =

    0
     
    Last edited: Aug 14, 2014
  11. Aug 14, 2014 #10

    LCKurtz

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    And the answer is yes. Have you tried what I suggested? It's much easier.
     
  12. Aug 14, 2014 #11

    etf

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    LCKurtz, I have done it using your substitution in only few lines. Definitely much better way to solve this integral :)
     
  13. Aug 14, 2014 #12

    etf

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    Can you tell me how did you come up with that equality?
     
  14. Aug 14, 2014 #13

    LCKurtz

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    Draw a picture of an angle whose tangent is ##z##. What is that angle's sine?
     
  15. Aug 14, 2014 #14

    etf

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    Got it now:)
     
  16. Aug 14, 2014 #15

    Ray Vickson

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    If we set x-1 = a you have
    [tex] \int \frac{y}{(a^2+y^2)^2} \, dy = \frac{1}{2a^2} \sin^2 \left( \arctan \left(\frac{y}{a} \right) \right)+C_1
    \equiv \frac{1}{2a^2} \frac{y^2}{a^2+y^2}+C_1 \: \Longleftarrow \: F_1(y)[/tex]
    Doing the integral another way gives
    [tex] \int \frac{y}{(a^2+y^2)^2} \, dy = -\frac{1}{2} \frac{1}{a^2+y^2} + C_2 \: \Longleftarrow \: F_2(y) [/tex]

    Although ##F_1(y)## and ##F_2(y)## look very different, it is wise to remember arbitrary constants of integration, so we have, in fact, that ##F_1(y) - F_2(y) = \text{const.}## Therefore, they are equivalent.
     
  17. Aug 16, 2014 #16

    mfb

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    Oh, right.
    Okay, looks fine.
     
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