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Math Help - help with integral

  1. #1
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    help with integral

    heres the problem: http://img186.imageshack.us/img186/4328/untitledil5.jpg

    i marked where i'm having trouble.

    I don't see how this was integrated.
    can someone break it down further, for the part i marked.

    thanks
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  2. #2
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    Quote Originally Posted by rcmango View Post
    heres the problem: http://img186.imageshack.us/img186/4328/untitledil5.jpg

    i marked where i'm having trouble.

    I don't see how this was integrated.
    can someone break it down further, for the part i marked.

    thanks
    The integral of 9/2 is 9/2 x.

    The integral of 6*sin x=-6cos x.

    The integral of 2*sin^2 x is tricker.
    You can write it using the half-angle identity to get,
    2*sin^2 x = 1-cos 2x and then that easily integrates.

    And the result is the sum of those.
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  3. #3
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    [QUOTE=1-cos 2x and then that easily integrates.QUOTE]

    so then using the trig identity, sinx^2 + cosx^2 = 1

    1 - cos2x is the integral.

    i'm going to guess what the derivative is, may need help, tried using mathematica.

    hmm not sure why x is dived by 2, wheres the 1/4 coming from? .. x/2 - 1/4sin(2x) + c

    thanks for all the help so far.
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  4. #4
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    so then using the trig identity, sinx^2 + cosx^2 = 1

    1 - cos2x is the integral.

    i'm going to guess what the derivative is, may need help, tried using mathematica.

    hmm not sure why x is dived by 2, wheres the 1/4 coming from? .. x/2 - 1/4sin(2x) + c

    thanks for all the help so far.
    The half-angle identity is:

    sin^2(x/2)=(1/2)(1-cos x)

    cos^2(x/2)=(1/2)(1+cos x)
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