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Math Help - Integral of cosine squared x?

  1. #1
    Member grgrsanjay's Avatar
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    Question Integral of cosine squared x?

    use integration by parts and let u=cosx and dv=cosxdx
    du=-sinx v=sinx
    Int(udv)=uv-Int(vdu) so uv=cosx(sinx) and vdu=sinx(-sinx)
    so we have:

    Int(cos^2(x)=(cosx)(sinx)+Int(sin^2x)
    (the (-) became + because of the -sinx, so we add Int(vdu))
    Now it looks not better because we have sin^2x instead of cos^2x,
    but sin^2x=1-cos^2x since sin^2x+cos^2x=1
    So we have

    Int(cos^2x)=cosxsinx+Int(1-cos^2x)
    =cosxsinx+Int(1)-Int(cos^2x)
    So now add the -Int(cos^2x) on the RHS to the one on the LHS
    2Int(cos^2x)=cosxsinx+x
    so Int(cos^2x)=1/2[cosxsinsx+x] and now add the constant!
    final answer

    Integral of cos^2x=(1/2)(cosx sinx + x)+C = x/2 + (1/4)sin 2x + C
    (because sin x cos x = (1/2)sin 2x)


    right??
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  2. #2
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    the answer is correct...
    but is there no other way...
    it is like round up.
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  3. #3
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    Simpler to use the trig identity cos^2(x)= \frac{1}{2}(1+ 2cos(x)) so that \int cos^2(x)dx= \frac{1}{2}\int dx+ \int cos(x)dx= \frac{1}{2}x+ sin(x)+ C.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Simpler to use the trig identity cos^2(x)= \frac{1}{2}(1+ 2cos(x)) so that \int cos^2(x)dx= \frac{1}{2}\int dx+ \int cos(x)dx= \frac{1}{2}x+ sin(x)+ C.

    i dont think ur answer is equal to cos^2(x)
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  5. #5
    Member grgrsanjay's Avatar
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    dint ever heard of it

    Quote Originally Posted by HallsofIvy View Post
    Simpler to use the trig identity cos^2(x)= \frac{1}{2}(1+ 2cos(x)) so that \int cos^2(x)dx= \frac{1}{2}\int dx+ \int cos(x)dx= \frac{1}{2}x+ sin(x)+ C.
    how did you get this identity??
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