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Math Help - More integration

  1. #1
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    More integration

    solve

    \int \frac{3+4\cos{x}}{(4+3\cos{x})^2}dx
    Last edited by mr fantastic; January 14th 2012 at 07:22 PM. Reason: Title.
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  2. #2
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    Re: integration (for more lulz...)

    Quote Originally Posted by Ozymandias View Post
    solve

    \int \frac{3+4\cos{x}}{(4+3\cos{x})^2}dx
    integral[(3 + 4*Cos[x])/((4 + 3*Cos[x])^2)] - Wolfram|Alpha

    Click "Show steps".
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  3. #3
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    Re: integration (for more lulz...)

    \begin{align*} \int \frac{3+4\cos{x}}{(4+3\cos{x})^2}dx &= \int \frac{3(\cos^2{x}+\sin^2{x})+4\cos{x}}{(4+3\cos{x}  )^2}dx \\ &= \int \frac{3\cos^2{x}+3\sin^2{x}+4\cos{x}}{(4+3\cos{x})  ^2}dx \\ &= \int \frac{\cos{x}(3\cos{x}+4)-\sin{x}(-3\sin{x})}{(4+3\cos{x})^2}dx \\ &=\int \frac{\frac{d(\sin{x})}{dx}\cdot(3\cos{x}+4)-\sin{x}\cdot\frac{d(3\cos{x}+4)}{dx}}{(4+3\cos{x})  ^2}dx \end{align*}

    Now recall that: \frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}=\frac{d \Big( \frac{f(x)}{g(x)}\Big)}{dx}

    Therefore the integral becomes: \int \frac{d \Big( \frac{\sin{x}}{4+3\cos{x}}\Big)}{dx}dx=\frac{\sin{  x}}{4+3\cos{x}}+C
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    Re: integration (for more lulz...)

    Thank You guyz, for your precious help...
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    Re: integration (for more lulz...)

    \displaystyle \int \frac{3+4\cos x}{(4+3\cos x)^2}dx

    divide both N_{r} and D_{r} by \sin^2 x

    \displaystyle \int \frac{3\csc^2 x+4\csc x.\cot x}{4\csc x+3\cot x}dx

    Now Put 4\csc x+3\cot x =t \Leftrightarrow -(4\csc x.\cot x+3\csc^2 x)dx = dt

    So \displaystyle -\int \frac{1}{t^2}dt = \frac{1}{t}+C

    So \displaystyle \int \frac{3+4\cos x}{(4+3\cos x)^2}dx = \frac{1}{4\csc x+3\cot x}+C
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