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Math Help - solve equation

  1. #1
    Junior Member
    Joined
    Jun 2008
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    48

    solve equation

    Hi everyone,

    I would like to know if I did any mistake in my calculation?

    Here is the question and my answer:

    (a Using equation:
    (f(x))n f(x) dx = 1/n+1(f(x))n+1 + c (n≠-1)


    show that
    cos x dx = - 1 + c,
    (2 + sin x)2 2+ sin x

    Where c is an arbitrary constant.

    cos x dx = - sin ∫ (x) dx
    (2 + sin x)2 (2+sinx)2

    = x = - 1
    (2+sin x)2 2+sin x + c



    Thanks for your help
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  2. #2
    Lord of certain Rings
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    IISc, Bangalore
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    Quote Originally Posted by valerie-petit View Post
    Hi everyone,

    I would like to know if I did any mistake in my calculation?

    Here is the question and my answer:

    (a Using equation:
    (f(x))n f(x) dx = 1/n+1(f(x))n+1 + c (n≠-1)


    show that
    cos x dx = - 1 + c,
    (2 + sin x)2 2+ sin x

    Where c is an arbitrary constant.

    cos x dx = - sin ∫ (x) dx
    (2 + sin x)2 (2+sinx)2

    = x = - 1
    (2+sin x)2 2+sin x + c



    Thanks for your help
    'Tis very hard to read what you have written. So I will work it out, check whether thats what you meant:

    \int \frac{\cos x}{(2 + \sin x)^2} \, dx

    If we choose f(x) =  2 + \sin x, then f'(x) = \cos x.

    Now

    \int \frac{\cos x}{(2 + \sin x)^2} \, dx = \int \frac{f'(x)}{f(x)^2}\, dx = \int f(x)^{-2} \, f'(x)\, dx = \frac{f(x)^{-1}}{-1}+C = -\frac{1}{2+ \sin x} + C
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  3. #3
    Junior Member
    Joined
    Jun 2008
    Posts
    48

    Thanks again

    Sorry about what I wrote, everytime I do copy/paste from word it does not recognize half of my text.

    Thanks again your help
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