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Math Help - integration by substitution

  1. #1
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    integration by substitution

    I'm completely lost with this.

    Need to calculate the integral \int {\frac{{\sin (\sqrt u )}}{{\sqrt u }}du} using the substitution  {\rm x  =  }\sqrt {\rm u}

    As far as I am getting is this
    <br />
\begin{array}{l}<br />
  = \int {\frac{{\sin (x)}}{x}du}  \\ <br />
  = \int {\frac{1}{x}\sin (x)du}  \\ <br />
 \end{array}<br />

    and that dx = \frac{1}{2}u^{\frac{{ - 1}}{2}} du

    Could someone solve and try and explain method. thanks
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  2. #2
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    Just sub z = \sqrt{u} You should get ∫2sin(z)
    Last edited by Chop Suey; October 5th 2008 at 09:12 PM. Reason: Correction
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  3. #3
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    Sorry but I still do not understand.
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  4. #4
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    \int \frac{\sin u^{\frac{1}{2}}}{u^{\frac{1}{2}}} \ du

    {\color{blue}x} ={\color{blue} u^{\frac{1}{2}}} \ \Rightarrow \ dx = \frac{1}{2}u^{-\frac{1}{2}} \ du  \iff {\color{red}2 dx} = {\color{red}\frac{1}{u^{\frac{1}{2}}} \ du}

    So: \int \frac{\sin {\color{blue}u^{\frac{1}{2}}}}{{\color{red}u^{\fra  c{1}{2}}}} \ {\color{red} du} = \int \sin {\color{blue}x} ({\color{red}2 \ dx}) = 2\int \sin x \ dx
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