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Integrate t.sin(t^2-1)

  1. Feb 5, 2012 #1

    sharks

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    The problem statement, all variables and given/known data

    Find [itex]\int t\sin(t^2-1)[/itex]

    The attempt at a solution
    I've used the method of integration by parts.
    Let U=t, then dU =1
    Let dV=[itex]sin(t^2-1)[/itex], then [itex]V=-\frac{cos(t^2-1)}{2t}[/itex]

    Then, [itex]\int t\sin(t^2-1)=t.-\frac{cos(t^2-1)}{2t}- \int -\frac{cos(t^2-1)}{2t}.1[/itex]

    Again, i use the method of integration by parts to find [itex]\int -\frac{cos(t^2-1)}{2t}.1[/itex]
    Let U=[itex]cos(t^2-1)[/itex], then dU=[itex]-\frac{sin(t^2-1)}{2t}[/itex]
    Let dV=[itex]\frac{-1}{2t}[/itex], then V=[itex]-\frac{ln2t}{2}[/itex]

    Then, [itex]\int -\frac{cos(t^2-1)}{2t}=cos(t^2-1).-\frac{ln2t}{2}-\int -\frac{ln2t}{2}.-\frac{sin(t^2-1)}{2t}[/itex]

    The integration cycle is endless. I'm stuck.

    Or, i could have also tried for [itex]\int -\frac{cos(t^2-1)}{2t}[/itex]
    Let [itex]U=\frac{-1}{2t}[/itex], then [itex]dU=\frac{2}{t^2}[/itex]
    Let [itex]dV=cos(t^2-1)[/itex], then [itex]V=\frac{sin(t^2-1)}{2t}[/itex]

    Then, [itex]\int -\frac{cos(t^2-1)}{2t}=\frac{-1}{2t}.\frac{sin(t^2-1)}{2t}-\int \frac{sin(t^2-1)}{2t}.\frac{2}{t^2}[/itex]

    But still, it looks like it's going to go on forever.
     
    Last edited: Feb 5, 2012
  2. jcsd
  3. Feb 5, 2012 #2
    Why not move the t into the differential. That way when you construct the equation you will have to find the derivative of the sine and then substitute...
    I will work that example out in a few mins
     
  4. Feb 5, 2012 #3

    sharks

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    I didn't understand what you meant, but i probably tried it already in my first post above. Anyway, i'm still waiting for your attempt at the solution.
     
  5. Feb 5, 2012 #4

    ehild

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    That is wrong.

    Use integration by substitution.


    ehild
     
  6. Feb 5, 2012 #5

    sharks

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    Hi ehild

    A tentative substitution... Let [itex]t=\sin\theta[/itex]. I tried it but it doesn't look "friendly". :smile:
    Here is what i get: [itex]\int \sin\theta\sin(-\cos^2\theta).\cos\theta\,.d\theta=\int \frac{\sin 2\theta}{2}.\sin(-\cos^2\theta)[/itex]
    I don't think i should use t=tan (half-angle) here. So, no idea what substitution to use.
     
    Last edited: Feb 5, 2012
  7. Feb 5, 2012 #6

    ehild

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    Why do you think it makes things easier???

    You have the composite function sin(t^2-1). It is multiplied by t which is almost the derivative of the inner function t^2-1. Choose the inner function as new variable.


    ehild
     
  8. Feb 5, 2012 #7

    sharks

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    Let [itex]u=t^2-1[/itex]

    [itex]t=\sqrt{u+1}[/itex]

    [itex]du=2t\,.dt[/itex], [itex]dt=\frac{1}{2\sqrt{u+1}}\,.du[/itex]

    Then, i get the answer: [itex]-\frac{\cos u}{2}=-\frac{\cos (t^2-1)}{2}[/itex]

    But in Wolfram, i get this answer: [itex]-\frac{\cos (1-t^2)}{2}[/itex]

    I suppose this means that the two answers are equivalent to each other?
     
    Last edited: Feb 5, 2012
  9. Feb 5, 2012 #8

    ehild

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    How are t^2-1 and 1-t^2 related?
    What about the parity of the cosine function?

    ehild
     
  10. Feb 5, 2012 #9

    sharks

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    [itex](t^2-1)=-(1-t^2)[/itex]

    [itex]\cos -(t^2-1)=\cos (t^2-1)[/itex]

    Therefore, [itex]-\frac{\cos (t^2-1)}{2}=-\frac{\cos -(1-t^2)}{2}=-\frac{\cos (1-t^2)}{2}[/itex]

    Am i correct?
     
  11. Feb 5, 2012 #10

    ehild

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    Well, you dropped a pair of parentheses. cos(-(t^2-1)) is correct.

    Is everything clear now? :smile:

    Do not forget that you can use substitution whenever you have have a factor in the integrand which is a constant times the derivative of an inner function.

    ehld
     
  12. Feb 5, 2012 #11
    Sharks, sorry... Got back from shopping - had to unload everything ...
    Now back to the problem: Indeed, this kind of substitution you wrote is the most efficient
    I tried solving by moving under the differential, but it gets too complicated and takes too much time to solve :)
    So, in conclusion - this is the way it should be done :)
     
  13. Feb 5, 2012 #12

    sharks

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    I understand it now. As usual, thanks for your help, ehild. :smile:
     
  14. Feb 5, 2012 #13
    The key to this is to use the double angle 2t. You need to convert the expression to be in terms of 2t instead of t squared.
     
  15. Feb 5, 2012 #14

    SammyS

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    In my opinion, some of the problems encountered in this thread arise from dropping the dt in the above integral, which should br written:
    [itex]\displaystyle \int t\sin(t^2-1)dt[/itex]​
    That way the substitution u = t2-1

    gives du=(2t)dt .

    Upon rewriting the integral it becomes pretty obvious where to plug everything in.

    [itex]\displaystyle \int t\sin(t^2-1)dt=\frac{1}{2}\int \sin(t^2-1)(2t)dt[/itex]
    [itex]\displaystyle =\frac{1}{2}\int \sin(u)du[/itex]​
     
  16. Feb 5, 2012 #15

    ehild

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    You are welcome (as usual :smile:) but I hope next time you remember the possibility of substitution.


    ehild
     
  17. Feb 5, 2012 #16

    ehild

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    You are right, Sammy, (as always :smile:) dt must not be omitted.

    ehild
     
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