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Integration problems. (Integration by parts)

  1. Feb 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Hi, I have a test coming up soon so I was doing some questions from the textbook when I stumbled upon this one and I'm stuck after like 5 tries. Here is the question:



    2. Relevant equations

    the question then states we should solve using this:
    cos^2(x)dx = (cos x)(cos x)

    which gives us:
    [tex]\int[/tex]cos^2(x)dx = sinxcosx + [tex]\int[/tex]sin^2(x)dx

    finally we should use sin^2(x) + cos^2(x) = 1 to replace the sin^2(x) at the right side of integral.

    3. The attempt at a solution
    so basically I tried using integral by parts, since we are studying this topic currently
    and set:

    u= 1 - cos^2(x) and dv = dx
    du = 2cosxsinx and v = x

    When I plug in the values into the integration and try to solve I don't end up with their answer. I must be overlooking something and this is where I am stuck:

    [tex]\int[/tex]cos^2(x)dx = sinxcosx + x(1-cos^2x) - [tex]\int[/tex]2xcosxsinx

    The final answer should be:
    [tex]\int[/tex]cos^2(x)dx = 1/2sinxcosx + 1/2x + C

    Thanks for the help.
  2. jcsd
  3. Feb 14, 2008 #2
    All problems of that type can be solved using this "formulas" :

    [tex]\int[/tex]cos^n(x)dx = 1/nsinxcos^n-1(x) +n-1/n [tex]\int[/tex]sin^n-2(x)dx

    [tex]\int[/tex]sin^n(x)dx = -1/nsin^n-1(x)cosx +n-1/n [tex]\int[/tex]sin^n-2(x)dx
  4. Feb 14, 2008 #3
    well, you could also use this identity

    cos^2(x)=(1+cos(2x))/2, it would really help you get to the answer pretty quickly, and without needing to do ineg by parts at all.
  5. Feb 14, 2008 #4


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    Science Advisor

    However, the point of the method that was suggested originally is that, after the first integration you have, as you say,
    [tex]\int cos^2(x)dx = sinxcosx + \int sin^2(x)dx[/tex]
    Now let [itex]sin^2(x)= 1- cos^2(x)[/itex] and that becomes
    [tex]\int cos^2(x)dx= sin(x) cos(x)+ \int (1- cos^2(x))dx[/tex]
    [tex]\int cos^2(x)dx= sin(x) cos(x)+ \int dx- \int cos^2(x) dx[/tex]
    [tex]\int cos^2(x)dx= sin(x) cos(x)+ x - \int cos^2(x) dx[/tex]
    Add [itex]\int cos^2(x)dx[/itex] to both sides of the equation and you are almost done.
    Last edited: May 12, 2009
  6. Feb 14, 2008 #5
    Halls off Ivy thanks a lot, I forgot that rule of integration. thanks a lot, yeah now that I tried it out it actually works. Thanks a lot.
  7. May 12, 2009 #6
    [tex] cos^2(x)dx= 1+cos2x/2[/tex]

    wat i wanna ask is.....is this the formula for cos^2x dx or can it be used wid [tex]\int[/tex] too??
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