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Integrate: Power of cos is even and non-negative

  1. Sep 2, 2007 #1
    First off, thanks for your help. I feel like a dope because everything I'm doing looks correct to me, but my answer is not the same as the book or solution manual gives! I'm usually so good at finding my mistakes.

    Thanks for taking the time to help.

    1. The problem statement, all variables and given/known data


    2. Relevant equations

    Where did I go wrong??

    3. The attempt at a solution


    The book's answer is -1/4 (cos x)^4 + C even when I try to convert sines to cosines, I'm not getting that those two answers were the same.
  2. jcsd
  3. Sep 2, 2007 #2


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    After the second step, try substituting [tex] u = cos (x) [/tex] instead. By the way, the two are equal, try substituting [tex] sin^2 (x) = (1-cos^2 (x)) [/tex] for every [tex] sin^2 (x) [/tex] in the answer and remember that [tex] \frac{1}{4}[/tex] is a constant as well.
    Last edited: Sep 2, 2007
  4. Sep 2, 2007 #3
    Thanks...I was able to get the book's answer, however...what is wrong with my solution?
  5. Sep 2, 2007 #4


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    Homework Helper

    Nothing looks wrong with your solution...i think your book just wanted you to immediately realize that u=cos(x) was a good substitution when you saw the integrand
  6. Sep 2, 2007 #5
    I have to go to work, I'll try the substitution again. Maybe just bad algebra trying to substitute. I'll post later if I got it or not.

  7. Sep 2, 2007 #6
    Substitute [tex]\sin^2{x} = 1 - \cos^2{x}[/tex] into your answer, and you should get the book's. Hence what you did wasn't wrong, just inefficient.
  8. Sep 3, 2007 #7
    Duh...got it....I think.

    Okay, thanks. It took a moment for it to sink in. When I substituted to try to get the two answers to equal up, I got [tex]\frac{1}{4} - [/tex] [tex]\frac{1}{4} cos^{4}x[/tex]

    I see now that the extra [tex]\frac{1}{4}[/tex] is just part of the constant of integration.

    If that is a correct statement, I am straight.

    Thank you! :approve:
  9. Sep 3, 2007 #8


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    By the way, you kind of confused me with your title: "Power of cos is even and non-negative". Obviously, here, both sine and cosine are to odd powers. The cosine substitution is the "standard" one for a situation like this.
  10. Sep 3, 2007 #9
    Thanks for the help..... It's all new to me!!!
    Last edited: Sep 3, 2007
  11. Sep 3, 2007 #10


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    Any time you have an integrand of the form sinn(x)cosm(x) with either m or n odd, you can always factor out 1 of those (for example if sin5(x)), so that you have left an even power (sin(x) sin4(x)). Use cos2(x)= 1- sin2(x) or sin2(x)= 1- cos2(x), repeatedly if necessary, to write it in terms of the other function. (sin4(x)= (sin2(x))2= (1- cos2(x))2 and then use the substitution u= cos(x) or u= sin(x).

    If both m and n are even then it is harder. Use sin2(x)= (1- 2sin(2x))/2 and cos2(x)= (1+ 2cos(2x))/2 to reduce the powers until you have an odd power of one or the other.
  12. Sep 3, 2007 #11
    Umm...you know what...you are absolutely right. Cos and sin are odd and positive. I work midnight shift and I posted it after not having sleep for quite some time.

    As for the even powers, I'm moving on to that homework tonight.

    Again, appreciate the help. This place is a gold mine of information!
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