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Thread: Integrating sin(sqrt x)?

  1. #1
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    Integrating sin(sqrt x)?

    I need to solve the problem:

    $\displaystyle \int\sin\sqrt x\ dx$

    Using integration by parts. So far, I've done:

    $\displaystyle u=\sqrt x$
    $\displaystyle du=\frac{1}{2\sqrt x}\ dx$
    $\displaystyle 2udu=dx$

    But then I'm lost.
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  2. #2
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    Hello, quarks!

    You're off to a good start . . .


    $\displaystyle \int\sin(\sqrt{x})\,dx$

    So far, I've done:

    $\displaystyle w\:=\:\sqrt x \quad\Rightarrow x \:=\:w^2\quad\Rightarrow\quad dx \:=\:2w\,dw$

    Substitute: .$\displaystyle \int 2w\sin w\,dw$

    Integrate by parts:
    . . $\displaystyle \begin{array}{ccccccc}u &=&2w & &dv &=&\sin w\,dw \\ du &=&2\,dw & & v &=&-\cos w\end{array}$

    And we have: .$\displaystyle -2w\cos w + 2\int\cos w\,dw \;=\;-2w\cos w + 2\sin w + C $

    Back-substitute: .$\displaystyle -2\sqrt{x}\cos(\sqrt{x}) + 2\sin(\sqrt{x}) + C$

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