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Math Help - Trig Integration by Parts Problem - # 4

  1. #1
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    Trig Integration by Parts Problem - # 4

    This one doesn't seem like an integration by parts problem.

    Integration by parts \int udv = uv - \int vdu

    \int 3\cos \sqrt{x} dx

    \int 3\cos x^{1/2} dx

    What is the next step? What is u? What is dv?
    Last edited by Jason76; July 2nd 2014 at 10:32 PM.
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  2. #2
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    Re: Trig Integration by Parts Problem - # 4

    Write it out like this:

    $\displaystyle \begin{align*} \int{ 3\cos{ \left( \sqrt{x} \right) } \, \mathrm{d}x} &= 6 \int{ \sqrt{x} \cos{ \left( \sqrt{x} \right) } \, \frac{1}{2\sqrt{x}}\,\mathrm{d}x } \end{align*}$

    Then make the substitution $\displaystyle \begin{align*} X = \sqrt{x} \implies \mathrm{d}X = \frac{1}{2\sqrt{x}}\,\mathrm{d}x \end{align*}$ and the integral becomes

    $\displaystyle \begin{align*} 6\int{ \sqrt{x}\cos{ \left( \sqrt{x} \right) } \, \frac{1}{2\sqrt{x}} \, \mathrm{d}x } &= 6 \int{ X \cos{(X)}\,\mathrm{d}X } \end{align*}$

    and now you can apply integration by parts.
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