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Integrating trig functions

  1. Jan 10, 2014 #1
    ## \frac {1}{4} \int sin^2 (2x)dx = I = \frac {1}{4} [- \frac {1}{2} sin(2x)cos(2x) + \int cos^2 (2x)dx]## when ##u = sin(2x), dv = sin(2x)dx, v= - \frac {cos(2x)}{2}## and ##du = 2cos(2x)dx##

    Now simplifying ##\int cos^2 (2x)dx## you get ## x - \int sin^2 (2x)dx = x - I##

    Then,

    ## I = \frac {1}{4} [- \frac {1}{2}sin(2x)cos(2x) + x - I)] ##

    ## \frac {5}{4} I = \frac {1}{4} [- \frac {1}{2}sin(2x)cos(2x) + x] ##

    ## I = \frac {1}{5} [ -\frac {1}{2}sin(2x)cos(2x) + x] ##

    Could anyone verify my solution? The next part is actually making this an indefinite integral of ##\int^{\frac {∏}{2}}_0 sin^2 (2x)dx## and when I solve this, my answer is always ##\frac {∏}{10}## and this is incorrect.
     
    Last edited: Jan 10, 2014
  2. jcsd
  3. Jan 10, 2014 #2
    I can't figure out where the 1/4 came from.
     
  4. Jan 10, 2014 #3

    vela

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    I didn't check your work carefully, but the final result doesn't look right. Did you try differentiating it and seeing if you recover the integrand?

    By the way, there's a much more straightforward way to do this integral than using integration by parts. Are you required to integrate by parts for this problem?
     
  5. Jan 10, 2014 #4
    Sorry, there was a ##\frac {1}{4}## in the beginning. Error on my part, sorry. I've corrected it in the original now.
     
  6. Jan 10, 2014 #5
    I tried it and got something different. I kept trying the question again but kept coming with the same integral. I understand I could have used half angle identities instead of integration by parts, but by using integration by parts. I keep getting an incorrect answer and if someone could point out what is wrong, that would be great.
     
  7. Jan 10, 2014 #6

    vela

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    The 1/5 is a problem. When I differentiated what's inside the square brackets, I got ##2\sin^2 2x##, so you're pretty close. It's probably just arithmetical errors somewhere.
     
  8. Jan 10, 2014 #7

    vela

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    The problem's here: ##I \ne \int \sin^2 2x\,dx##
     
  9. Jan 10, 2014 #8
    ##I = \frac {1}{4}\int \sin^2 2x\,dx## is the correct equation, right?

    Okay, so:

    ## \frac {1}{4}I = \frac {1}{4} [- \frac {1}{2}sin(2x)cos(2x) + x - I)] ##

    ## \frac {1}{2} I = \frac {1}{4} [- \frac {1}{2}sin(2x)cos(2x) + x] ##

    ## I = \frac {1}{2} [ -\frac {1}{2}sin(2x)cos(2x) + x] ##

    In case I didn't do exactly what you wanted, I still get ##\frac {∏}{4}## as the answer which is still not right...
     
  10. Jan 10, 2014 #9
    Have you tried using the identity [itex]sin^2θ=\frac{1-cos(2θ)}{2}[/itex]?
     
  11. Jan 10, 2014 #10

    vela

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    Yes, you defined it that way, after all. :wink: So ##\int \sin^2 2x\,dx = 4I##, right?

    It would be better to say that
    $$I = \frac{1}{4}\left[x-\frac{1}{2}\sin 2x\cos 2x - 4I\right]$$ rather than changing the definition of ##I## midstream, which is probably why you're getting the wrong answer.

     
  12. Jan 10, 2014 #11
    Also, instead of making new threads, I have a related question.

    Given:

    ## \int sin(x)cos(x)dx = I ## and ##u = sin(x), du = cos(x)dx## so ## I = \int udu = \frac {1}{2}sin^2(x)##

    Using ##w = cos(x), -dw = sin(x)dx## I found ## I = -\int wdw = -\frac{1}{2}w^2 = -\frac{1}{2}cos^2(x)##

    Using the two methods, I found two different answers and I am not completely sure why....
    Once again, and clarification on what I did wrong or am thinking about incorrectly would be very helpful! :)
     
  13. Jan 10, 2014 #12

    vela

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    Those two results differ only by a constant.
     
  14. Jan 10, 2014 #13
    Oh boy...let's not make that mistake again haha thank you, I should really develop better strategies to check my work.
     
  15. Jan 10, 2014 #14
    [tex] \frac {1}{4} \int sin^2 (2x)dx = \frac {1}{32} \int {(1-cos(4x))d(4x)}=\frac{x}{8}-\frac{sin(4x)}{32}+C[/tex]
     
  16. Jan 10, 2014 #15
    And there we go...defining ## I ## early and changing it was a bad idea...Thank you everyone for the great help!
     
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