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Product rule in differentiation help

  1. Apr 14, 2005 #1
    [tex] \int x sin (2x) dx =?[/tex]
    [tex]u=2x[/tex]
    [tex]du=2dx[/tex]
    [tex]\int \frac{u}{2} sin(u) \frac{du}{2}[/tex]
    [tex]\frac{1}{4} \int usin(u)du[/tex]
    [tex]\frac{1}{4} (\frac{u^2}{2}sin(u)+u(-cos(u)))[/tex]
    [tex]\frac{2x^2}{4} sin(2x)-\frac{1}{2} cos(2x)+C[/tex]

    is this correct?
     
    Last edited: Apr 14, 2005
  2. jcsd
  3. Apr 14, 2005 #2

    dextercioby

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    Nope,it's incorrect.U should have part integrated directly

    [tex] \int x\sin 2x \ dx=-\frac{1}{2}x\cos 2x+\frac{1}{2}\int \cos 2x \ dx=-\frac{1}{2}x\cos 2x+\frac{1}{4}\sin 2x+\mathcal{C} [/tex]


    Daniel.
     
  4. Apr 14, 2005 #3
    Try parts...
     
  5. Apr 14, 2005 #4
    Opps I did not see that you had already posted...
     
  6. Apr 14, 2005 #5
    I haven't learned parts yet, I dont not follow your process
     
  7. Apr 14, 2005 #6
    You can't solve that integral by substitution.
     
  8. Apr 14, 2005 #7
    By parts you set

    u = x, du = dx, v = -cos(2x)/2, dv = sin(2x)dx then

    [tex] \int u dv = uv - \int v du [/tex] which follows from the product rule of differentiation.

    [tex] \int xsin(2x)dx = \frac{-xcos(2x)}{2} + \frac{1}{2}\int{cos(2x)dx} [/tex]

    [tex] \int{xsin(2x)dx} = \frac{-1}{2}xcos(2x) + \frac{1}{4}sin(2x)[/tex]
     
  9. Apr 14, 2005 #8
    what exactly is parts and is there a website that explains it?

    why cant I solve it with subsitution?
     
  10. Apr 14, 2005 #9
    Integrating by parts is just taking advantage of the result of the using the product rule in differentiation. See here. You choose a factor to integrate and a factor to differentiate in order to make the integral which appears in the right-hand side of the parts equation easier than the integral you started with.
    I doubt you can solve the above by substitution alone, as it involves a product of elementary functions, no matter what substitution you make.
     
    Last edited: Apr 14, 2005
  11. Apr 14, 2005 #10
    so the basic rule to integration by parts is:

    [tex]\int u dv= uv- \int v du[/tex]
    correct?
     
    Last edited: Apr 14, 2005
  12. Apr 14, 2005 #11
    Yep. Choose u and dv such that v*du is a simpler integral. In your case, I would let u = x and v = sin(2x) dx, so that du = dx and v = -(1/2)*cos(2x).
     
    Last edited: Apr 14, 2005
  13. Apr 14, 2005 #12

    dextercioby

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    U can do it using a substitution & power series,too_Or only power series...

    Daniel.
     
  14. Apr 14, 2005 #13
    so I use parts usually when I'm trying to integrate xsin(x) or xcos(x) or xtan(x) ?

    what is power series?
     
  15. Apr 14, 2005 #14

    dextercioby

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    It's quite difficult to integrate

    [tex] I=:\int x\tan x \ dx [/tex] (1)

    ,if u don't know some tricks.

    Assume [itex]\cos x\neq 0 [/itex] (2)

    Use parts

    [tex] \int x\tan x \ dx =-x\ln\cos x+\int \ln\cos x \ dx [/tex] (3)

    Denote

    [tex] J=:\int \ln\cos x \ dx [/tex] (4)

    Now,u know that

    [tex] \cos x=\frac{e^{ix}+e^{-ix}}{2} [/tex] (5)

    Therefore

    [tex] J=\int \ln\left(\frac{e^{ix}+e^{-ix}}{2}\right) \ dx [/tex] (6)

    Use the property of the logarithm to write

    [tex] J=\int \ln\left[e^{ix}\left(1+e^{-2ix}\right)\right] \ dx -\int \ln 2 \ dx =\int \ln e^{ix} \ dx +\int \ln\left(1+e^{-2ix}\right) \ dx - x\ln 2 [/tex]
    [tex]= i\frac{x^{2}}{2}-x\ln 2 + K [/tex](7)

    ,where

    [tex] K=:\int \ln\left(1+e^{-2ix}\right) \ dx [/tex] (8)

    Make the substitution

    [tex] e^{-2ix}=u [/tex] (9)

    [tex] dx=\frac{i}{2u} du [/tex] (10)

    Then K becomes

    [tex] K=\frac{i}{2}\int \left[\frac{\ln(1+u)}{u}\right] \ du =-\frac{i}{2}\mbox{dilog}(1+u) +\mathcal{C}=-\frac{i}{2}\mbox{dilog}\left(1+e^{-2ix}\right)+\mathcal{C} [/tex] (11)

    Therefore,from (1),(3),(4),(7),(11),one gets

    [tex] \int x\tan x \ dx=-x\ln\cos x+i\frac{x^{2}}{2}-x\ln 2-\frac{i}{2}\mbox{dilog}\left(1+e^{-2ix}\right)+\mathcal{C} [/tex]

    Daniel.
     
    Last edited: Apr 14, 2005
  16. Apr 14, 2005 #15

    dextercioby

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    Power series:series expansions (Taylor,Laurent) of analytical functions.It's a method to integrate.However,care must be taken with the convergence of such series...


    Daniel.
     
  17. Apr 15, 2005 #16
    I did not know that [tex] \cos x=\frac{e^{ix}+e^{-ix}}{2} [/tex]

    what is the expression [tex]e^{ix}[/tex] mean?
     
  18. Apr 15, 2005 #17

    dextercioby

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    The famous Euler's formula

    [tex] e^{ix}=\cos x+i\sin x [/tex]


    Daniel.
     
  19. Apr 15, 2005 #18

    HallsofIvy

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    it means, of course, e to the power ix where i is the "imaginary unit": i2= -1.

    One problem I have with your posts is that you seem to be posting problems well well in advance of what you know. That is, you post calculus problems that require some basic algebra technique and when someone points that out, you don't know the algebra. I would expect that someone who was taking calculus would have already been exposed to imaginary numbers and, even if not, comfortable with them, have seen eix= cos(x)+ i sin(x) before.
     
  20. Apr 15, 2005 #19
    I guess there is some lack of knowledge on my part then. Probably my basic foundation of math has not been ingrained in my brain yet, that's why I might ask some basic questions to be explained, just to reinforce these foundations. However, I am greatly in debt for all the help and explanations that I have gotten, so thank you all helping me on my 800+ posts :smile:

    by the way, I have never seen eix= cos(x)+ i sin(x) before
     
  21. Apr 15, 2005 #20
    In all fairness I've completed the calc requirement for physics and only saw that equation after coming to PF.

    [tex] e^{ix}= cos(x)+ i sin(x) [/tex]

    Though ofcourse I immediately looked it up and found the derivation and reasoning for it, so I'm cool now. :)
     
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