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Indefinite Integral - II

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \left(\frac{x}{x\cos x-\sin x} \right)^2 dx[/tex]


    2. Relevant equations



    3. The attempt at a solution
    Factoring out ##\cos x## from the denominator, the integral transforms to
    [tex]\int \sec^2x \left(\frac{x}{x-\tan x}\right)^2dx[/tex]
    Substituting ##\tan x=t##, ##\sec^2 xdx=dt##
    [tex]\int \left(\frac{\arctan t}{\arctan t-t}\right)^2dt[/tex]
    I honestly have no idea about how to proceed from here.

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Apr 28, 2013 #2

    SteamKing

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    I don't think you can factor cos x from sin x and obtain tan x. I suggest you expand the terms within the brackets and see if it can be simplified with trig identities.
     
  4. Apr 28, 2013 #3
    Why can't I factor out cosx?
    sinx/cosx=tanx
     
  5. Apr 28, 2013 #4

    SteamKing

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    The problem is you have only sin x in your integral.
     
  6. Apr 28, 2013 #5

    haruspex

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    Look at the derivative of x cos(x) - sin(x). This suggests a way to use integration by parts.
     
  7. Apr 28, 2013 #6
    The derivative of xcos(x)-sin(x) is -xsin(x) but I still can't figure it out. The whole thing is squared, I am completely lost.
     
  8. Apr 28, 2013 #7

    SteamKing

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    You could have expanded the brackets by now and checked to see if trig identities could further simplify the integrand.
     
  9. Apr 28, 2013 #8

    haruspex

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    It suggested to me writing the integrand as ##\frac{x}{\sin x} \frac{x\sin x}{\left(x\cos x-\sin x \right)^2} ##. Try integrating that by parts.
     
  10. Apr 29, 2013 #9
    Integrating by parts:
    [tex]\frac{x}{\sin x} \int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx-\int \frac{\sin x-x\cos x}{\sin^2 x} \left(\int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx \right)dx[/tex]

    [tex]\because \int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx=(x\cos x-\sin x)[/tex]
    The integral simplifies to:
    [tex]\frac{x}{\sin x}(x\cos x-\sin x)+\int \left(\frac{x \cos x-\sin x}{\sin x} \right)^2dx[/tex]
    If I expand the squared term, I get an integral involving cot and cot^2, should I integrate by parts again?
     
  11. Apr 29, 2013 #10

    haruspex

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    Retry that step.
     
  12. Apr 29, 2013 #11

    SammyS

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    [STRIKE]How is that integration by parts? What were the parts you used?[/STRIKE]

    Try ##\displaystyle \ u=\frac{x}{\sin(x)}\ \text{ and }\ dv=\frac{x\sin(x)}{\left(x\cos(x)-\sin(x) \right)^2}\,dx \ . ##

    Added in Edit: Part of the above edited out.
    I see it now .
     
    Last edited: Apr 29, 2013
  13. Apr 29, 2013 #12

    haruspex

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    Looks to me that Pranav-Arora was getting there - just made a mistake in the first integration.
     
  14. Apr 29, 2013 #13

    SammyS

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    Oh ! I see it now.

    I had looked at it several times and didn't see how that fit integration by parts.
     
  15. Apr 30, 2013 #14
    Oops, its the reciprocal of that.

    This time I get:
    [tex]\frac{x}{\sin x}\frac{1}{x\cos x-\sin x}+\int \csc^2 xdx=\frac{x}{\sin x}\frac{1}{x\cos x-\sin x}-\cot x[/tex]

    But on further simplification, I do not end up with the right answer.
     
  16. Apr 30, 2013 #15

    ehild

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    It is correct, bring to common denominator and simplify with sinx.Show what you did.

    ehild
     
  17. Apr 30, 2013 #16
    Got it, thanks! :smile:

    Writing cot in terms cos and sin
    [tex]\frac{x-x\cos^2 x+\cos x \sin x}{\sin x(x\cos x-\sin x)}=\frac{x(1-\cos^2 x)+\cos x \sin x}{\sin x(x\cos x-\sin x)}[/tex]
    As 1-cos^2x=sin^2x
    [tex]\frac{x\sin^2 x+\cos x \sin x}{\sin x(x\cos x-\sin x)}=\frac{x\sin x+\cos x }{x\cos x-\sin x}[/tex]

    Thanks a lot haruspex! :smile:
     
    Last edited: Apr 30, 2013
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