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Math Help - Integral of cos^2(x)dx

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    26

    Integral of cos^2(x)dx

    Problem:

    \int cos^2(x)dx

    Attempt:

    u = cos(x)
    du = -sin(x)dx
    dv = cos(x) dx
    v = sin(x)
    cos(x)sin(x) - \int -sin(x)sin(x)

    cos(x)sin(x) + \int sin(x)sin(x)

    Focusing on \int sin(x)sin(x)
    u = sin(x)
    du = cos(x)dx
    dv = sin(x)
    v = -cos(x)
     <br /> <br />
-sin(x)cos(x) - \int cos(x)-cos(x)<br />

    -sin(x)cos(x) + \int cos(x)cos(x)

    Am I doing this wrong?

    Correct Answer:
    \frac{sin(2x)}{4} + x/2 + C
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  2. #2
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    Your answer is correct. Though it would be easier to just use the fact that \cos^2 x = \frac{1+\cos 2x}{2} and \sin^2 x = \frac{1-\cos 2x}{2}
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  3. #3
    Senior Member chella182's Avatar
    Joined
    Jan 2008
    Posts
    267
    Quote Originally Posted by Spec View Post
    Your answer is correct. Though it would be easier to just use the fact that \cos^2 x = \frac{1+\cos 2x}{2}
    Yeah, this is derived from the double angle formula \cos{2x}=2\cos^2x-1
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