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Integral of complicated trig function.

  1. Jul 24, 2011 #1
    1. The problem statement, all variables and given/known data
    This question has me stumped Please help me get this monster completed.
    Compute the definite integral of -2011 to 2011 of (1+sin^2(17t))^2011*sin(sin(-t))dt

    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 25, 2011 #2
    Re: Desperate Help!!!!

    I think you can use symmetry to help you. Is the function odd?
     
  4. Jul 25, 2011 #3
    Re: Desperate Help!!!!

    Replace 2011 with n. Use integration by parts to get a recurrent relation I(n) = F(I(n-1), I(n-2)...). Use this recurrent relation to find the general value for I(n) and afterthat replace n with 2011.
     
  5. Jul 25, 2011 #4

    BruceW

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    Re: Desperate Help!!!!

    atomthick has the right idea.
    To explain it in my words (to make sure you understand what to do):
    When you use the product rule a couple of times, (differentiating the sin to the power bit), hopefully you will see a pattern in how the function changes each time you use product rule. So then you make an equation that gives the total change to the function after n times. then use n=2010, which will effectively get rid of that power 2011, so then you can try using normal method of integration.
    (I haven't actually tried this specific problem, but this is generally what I would try doing).
     
  6. Jul 25, 2011 #5

    BruceW

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    Re: Desperate Help!!!!

    The function looks even.
    Actually, dcee27's notation is a bit confusing. In one place he has sin^2(17t) and in another place he has sin(sin(-t)).
     
  7. Jul 25, 2011 #6
    Re: Desperate Help!!!!

    spamiam has seen the easy way out, indeed the function is odd.

    F(t) = -F(-t) and F(0) = 0.
     
    Last edited: Jul 25, 2011
  8. Jul 25, 2011 #7

    BruceW

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    Re: Desperate Help!!!!

    Why is it odd?
    [tex] (1 + sin^2(17t) )^{2011} \ sin^2(-t) [/tex]
    Looks like an even function to me.
     
  9. Jul 25, 2011 #8
    Re: Desperate Help!!!!

    Thanks Robert1986 but I'm not an atomCHICK :))
     
  10. Jul 25, 2011 #9
    Re: Desperate Help!!!!

    I belive the formula is actualy [tex] (1 + sin(17t)\sin(17t) )^{2011} \ sin(sin(-t)) [/tex] but dcee27 knows better...
     
  11. Jul 25, 2011 #10

    BruceW

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    Re: Desperate Help!!!!

    Woops, yeah I got sin squared and sin of sin mixed up.
     
  12. Jul 25, 2011 #11
    Re: Desperate Help!!!!

    The question is entered correclty. (1+sin^2(17t))^2011 * sin(sin(-t))dt bounded by a = -2011 and b= 2011 o fthe definite intergral.

    I still can't compute!!!
     
  13. Jul 25, 2011 #12

    micromass

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    Re: Desperate Help!!!!

    Can you show that the function is odd??
     
  14. Jul 26, 2011 #13
    Re: Desperate Help!!!!

    Yes, this part b of a previous question that asked to prove that the a integral was an odd function. So I would say yes.
     
  15. Jul 26, 2011 #14

    micromass

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    Re: Desperate Help!!!!

    So, what is the integral of an odd function?
     
  16. Jul 26, 2011 #15
    Re: Desperate Help!!!!

    the first part asked to explain and show that why if f(x0 is an odd function integral a= -a and b = a f(t)dt = 0. That is part a and then part b which is posted asked to compute and explain the integral i posted.
     
  17. Jul 26, 2011 #16
    Re: Desperate Help!!!!

    OK, I want you to do two things:

    1) Draw the graph from -5 to 5 of some odd function (in fact, just do a few, sin(x), x^x, x)

    2) Look at this graph.

    3) What is the area under this curve from -5 to 5 (remember, area under the x axis is "negative area".)

    4) Now use what you have done and see if you can generalise to ANY odd function from -a to a.
     
  18. Jul 26, 2011 #17

    BruceW

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    Re: Desperate Help!!!!

    So, part a) says that the integral of an odd function from -a to a equals zero.
    And you've said that the function you were given for part b) is an odd function.
    So simply use the fact from part a) to find out the answer to b) (its simpler than you think).
     
  19. Jul 26, 2011 #18
    Re: Desperate Help!!!!

    There are three types of people in the world: those who can count and those who can't.

    Its good to see I belong in the first. :)
     
  20. Jul 27, 2011 #19
    Re: Desperate Help!!!!

    Erm, x^x isn't an odd function either... :uhh:
     
  21. Jul 27, 2011 #20
    Re: Desperate Help!!!!

    I'm not sure what I mean by x^x; I'm guessing I meant x^2 or something.
     
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