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Math Help - integration by parts of sin^2 of x

  1. #1
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    integration by parts of sin^2 of x

    I know that I can use an identity to solve this but I am suppose to use integration by parts and I can not figure out how!!! Any advice would be much appreciated. Thank you
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  2. #2
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    Quote Originally Posted by Frostking View Post
    I know that I can use an identity to solve this but I am suppose to use integration by parts and I can not figure out how!!! Any advice would be much appreciated. Thank you
    Let  I = \int \sin^2 x \, dx.

    Let u = \sin x and dv = \sin x \, dx:

    I = -\sin x \cos x + \int \cos^2 x \, dx = -\sin x \cos x + \int dx - I.

    Now solve for I by making it the subject.
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  3. #3
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    integral sin^2 x by parts

    I think that I must be the stupidest person alive but I can not get this.


    I get sin(x)(-cos(x) - integral of (-cos(x) times cos(x) dx

    which goes to sin(x)(-cos(x) + integral of cos^2(x) dx

    and now I have the same type of problem all over again and I do not see how using by parts will be anything but an endless loop!!! I know I am missing the point but I feel helpless and hopeless to figure out why. Can anyone explain it to me?????
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    Let  I = \int \sin^2 x \, dx.

    Let u = \sin x and dv = \sin x \, dx:

    I = -\sin x \cos x + \int \cos^2 x \, dx = -\sin x \cos x + \int dx - I.

    Now solve for I by making it the subject.
    Quote Originally Posted by Frostking View Post
    I think that I must be the stupidest person alive but I can not get this.


    I get sin(x)(-cos(x) - integral of (-cos(x) times cos(x) dx

    which goes to sin(x)(-cos(x) + integral of cos^2(x) dx

    and now I have the same type of problem all over again and I do not see how using by parts will be anything but an endless loop!!! I know I am missing the point but I feel helpless and hopeless to figure out why. Can anyone explain it to me?????
    I = -\sin x \cos x + \int \cos^2 x \, dx

    {\color{red}=  -\sin x \cos x + \int (1 - \sin^2 x) \, dx = -\sin x \cos x + \int 1 \, dx - \int \sin^2 x \, dx}

    = -\sin x \cos x + \int dx - I

    {\color{red} \Rightarrow 2I = -\sin x \cos x + \int dx = -\sin x \cos x + x + C}

    {\color{red} \Rightarrow I = -\frac{1}{2} \sin x \cos x + \frac{x}{2} + K}.
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