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Math Help - [SOLVED] integrating factor gone wrong

  1. #1
    Member i_zz_y_ill's Avatar
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    [SOLVED] integrating factor gone wrong

    ok so i dont get this question, the integrating factor is clearly sinx from which point i decided that if you minus each side you can abreviate it to  \frac{d(-sinxf)}{dx}=-cos^2xsinx NB original question is: \frac{df}{dx}+fcotx=cos^2x the answer is somehow f(x)=-\frac{cos^3x}{3sinx}+\frac{C}{sinx} which i do no understand,,,how can the constand have a sinx underneath is isnt this just another constant? I must be going wrong somewhere on the integral of \int cos^2dx which i make out to be,,, \frac{co3^3x}{3} ? any clues?
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  2. #2
    Senior Member Spec's Avatar
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    \frac{d}{dx}\left(\frac{1}{3}\cos^3 x \right)\neq \cos^2 x
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  3. #3
    Member i_zz_y_ill's Avatar
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    oh yeah! now using \int \frac{1}{2}+\frac{cos2x}{2}dx= ???
    still cant get there answer
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  4. #4
    Super Member Random Variable's Avatar
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     sin(x) \frac {df}{dx} + f sin(x) \frac {cosx}{sinx} = sin(x)cos^{2}x

     \int \frac {d}{dx} (fsin(x))dx = \int sin(x)cos^{2}(x)dx

     fsin(x) = - \int u^2 du = -\frac {1}{3}u^{3} + C = - \frac {1}{3}cos^{3}x + C

    then solve for f
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  5. #5
    Member i_zz_y_ill's Avatar
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    what is u? still bit unsure how to solve that last integral?  \int \frac{sinx}{2}+\int sinx -2 \int sin^3x ?

     sinxcos^2x=sinx(\frac{1}{2}+cos2x)=\frac{sinx}{2}+  sinx(1-2sin^2x)?

    any ideas?
    Last edited by Krizalid; May 29th 2009 at 07:13 PM.
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  6. #6
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     \int \sin x \cos^2 x dx =\left\{u=\cos x, du=-\sin x dx \implies dx=-\frac{du}{\sin x}\right\}=\int -u^2 du
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