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Thread: integration by substitution

  1. #1
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    integration by substitution

    I'm completely lost with this.

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

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

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

    Could someone solve and try and explain method. thanks
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  2. #2
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    Just sub $\displaystyle z = \sqrt{u}$ You should get ∫2sin(z)
    Last edited by Chop Suey; Oct 5th 2008 at 08:12 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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    $\displaystyle \int \frac{\sin u^{\frac{1}{2}}}{u^{\frac{1}{2}}} \ du$

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

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