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Integration by parts with trig function insanity

  1. Feb 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Let In=the integral from 0 to pi/2 of sinnxdx
    show that (I2n+1)/(I2n)=(2n+1)/(2n+2)


    2. Relevant equations
    integral from 0 to pi/2 of sin2n=(2n-1)pi/4n


    3. The attempt at a solution
    I can't write down all that I've done because it's just too ridiculous. I've tried lots of forms of just trying to manipulate my equation above, which didn't work. I've tried integration by parts for sin2n+1*sinx but I've ended up with the integral of cos2xsin2nx and I can't find a suitable substitution for that. I've been working on this problem forever and it's killing me. If anyone has any suggestions I would be very appreciative.
     
  2. jcsd
  3. Feb 19, 2009 #2
    Have you seen the derivation of Wallis' infinite product for pi/2?
     
  4. Feb 19, 2009 #3
    unfortunately, I have to use this problem to prove that next.
     
  5. Feb 21, 2009 #4

    lanedance

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    Homework Helper

    so you're trying to do the even integral first? think you might have been on the right track how about looking at some induction as a next step...

    I2n = [tex]\int[/tex]dt sint2n = [tex]\int[/tex]dt sint2n-1.sint

    integrating by parts

    I2n = (2n-1)[tex]\int[/tex]dt sint2(n-1)cost2

    I2n = (2n-1)[tex]\int[/tex]dt sint2(n-1) - (2n-1)[tex]\int[/tex]dt sint2n

    So
    2n. I2n = (2n-1).I2(n-1)
     
  6. Feb 21, 2009 #5

    lanedance

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    Homework Helper

    so you're trying to do the even integral? think you might have been on the right track how about looking at some induction as a next step...

    so from I2n = [tex]\int[/tex]dt sint2n = [tex]\int[/tex]dt sint2n-1.sint

    and integrating by parts like you did, I get to

    2n. I2n = (2n-1).I2(n-1)

    Also is that formula for I2n in the post correct? i would of thought it should tend zero as n heads towrds infinity
     
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