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Integration of 1/2x & sinxcosx & x+1

  1. May 5, 2010 #1
    Hello I was wondering why in this differential equation they did not use double angle identity, I understand how to get the other result also


    1)Dif eqn:

    Dy/dx +ycotx= cosx

    Integrating factor IF= e^ (integrand of cotx) = sinx

    yIF= integrand of QIF where Q is cotx from original question)

    y sinx= integral of cosxsinx

    Now I recognised this as ½ sin2x and the integral would be

    ysinx= ½(-cos2x/2+ c)

    therefore y= 1/sinx (-cos2x/4+ c/2)

    however this could be done by substitution

    y sinx= integral of cosxsinx let u=sinx to give
    ysinx= integral of cosx./u du/cosx
    ysinx= sin^2x +c
    y=1/sinx (sin^2x +c)

    why are they not the same ?? or are they really the same

    I mean if we differentiate either sin^2x or -cos2x/2 we would get back to cosxsinx

    (x-2)dy/dx –y = (x-2)^3
    dy/dx – y/(x-2)= (x-2)^2
    IF= 1/(x-2)
    yIF= intergral of QIF

    y/(x-2)= integral of (x-2)^2 .1/(x-2)
    y/(x-2)= integral of (x-2)

    If I were to integrate (x-2) why is it (x-2)^2/2 +c is it the same as x^2/2-2x+c
    (I am confused about this one as from expansion I can see its similar but if I put any value for c into either say 1 I get different answers)

    3)Finally integral of 1/2x I can do this by dividing by a half
    i.e ½ integral of 1/x =1/2 lnx

    but is itthe same as 1/2ln2x as this wway I say what must I do to denominator (2x) to get to numerator and then simply multiply this coefficient to the natural log of the bottom

    i.e 1/4x+3 = ½ ln 4x+3

    Can I only do that if I cannot factorise out a coeff?
  2. jcsd
  3. May 5, 2010 #2

    D H

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    Staff Emeritus
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    Two typical reasons why using two integration techniques yield different answers are
    • You made a mistake in one or both integrations.
    • The answers are the same, to within an arbitrary constant.

    You did both here. [itex]\int u du = 1/2u^2 + c[/itex]. You forgot the factor of 1/2.

    The trig substitution and u substitution techniques yield solutions of

    y\sin x &= -\,\frac 1 4 \cos 2x + c_1 && \text{trig substitution} \\
    &= \phantom{-}\frac 1 2 \sin^2 x + c_2 && \text{ u-substitution}

    With the trig substitution [itex]\cos 2x = 1 - 2\sin^2x[/itex], the first becomes

    [tex]y\sin x = -\,\frac 1 4 (1-2\sin^2 x) + c_1 = \frac 1 2 \sin^2 x + c_1 - \frac 1 4[/itex]

    So the two techniques do agree to within an arbitrary constant.
  4. May 5, 2010 #3
    so a diff technique will give a diff constant?
  5. May 5, 2010 #4

    D H

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    Staff Emeritus
    Science Advisor

    Not always, but in this case, yes.
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