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Thread: Integration by Parts

  1. #1
    Newbie Mr. Engineer's Avatar
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    Integration by Parts

    Hello, I just have two problems. I have used integration by parts, but I get stuck. The work you show will be greatly appreciated.

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

    $\displaystyle \int e^{6x} sin(e^{2x})~dx$

    Thanks!
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  2. #2
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    Quote Originally Posted by Mr. Engineer View Post
    Hello, I just have two problems. I have used integration by parts, but I get stuck. The work you show will be greatly appreciated.

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

    $\displaystyle \int e^{6x} sin(e^{2x})~dx$

    Thanks!
    For the first one, do you mean $\displaystyle \sqrt{x} \sin(x) $ ??
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  3. #3
    Newbie Mr. Engineer's Avatar
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    no, its the way it is, i trued just using u-substitution, but the answer was not what I got, we were supposed to use integration by parts to solve that one
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    Quote Originally Posted by Mr. Engineer View Post
    Hello, I just have two problems. I have used integration by parts, but I get stuck. The work you show will be greatly appreciated.

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

    $\displaystyle \int e^{6x} sin(e^{2x})~dx$

    Thanks!
    Well, what are you waiting for? Do it!

    $\displaystyle
    \int{sin(\sqrt{x})}\;dx\;=\;x \cdot sin(\sqrt{x})\;-\;\int x \;d\left(sin (\sqrt{x}) \right)\;=\;x \cdot sin(\sqrt{x})\;-\;\int{x \cdot \frac{cos(\sqrt{x})}{2 \cdot \sqrt{x}}}\;dx
    $

    Now what?
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  5. #5
    Eater of Worlds
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    Do you have to use parts?. I think a sub is easier.

    $\displaystyle \int e^{6x}sin(e^{2x})dx$

    Now, let $\displaystyle u=e^{2x}, \;\ \frac{1}{2}du=e^{2x}dx$

    Make the subs and we get:

    $\displaystyle \frac{1}{2}\int u^{2}sin(u)du$

    Now, let's use tabular integration:

    This works well when we have a something such as e or sin that repeats as we take derivatives. It's based off parts, though. So, I reckon we are using parts.

    $\displaystyle \text{\underline{sign}}, \;\ \;\ \text{\underline{u and its derivatives}} \;\ \text{\underline{v' and its antiderivatives}}$

    $\displaystyle + \;\ \;\ \;\ \;\ \rightarrow \;\ \;\ \;\ \;\ u^{2} \;\ \;\ \;\ \;\ \searrow \;\ \;\ \;\ \;\ \;\ \;\ \;\ v'=sin(u)$

    $\displaystyle - \;\ \;\ \;\ \;\ \;\ \rightarrow \;\ \;\ \;\ \;\ 2u, \;\ \;\ \;\ \;\ \searrow \;\ \;\ \;\ \;\ \;\ -cos(u)$

    $\displaystyle + \;\ \;\ \;\ \;\ \;\ \rightarrow \;\ \;\ \;\ \;\ 2 \;\ \;\ \;\ \;\ \searrow \;\ \;\ \;\ \;\ \;\ \;\ -sin(u)$

    $\displaystyle - \;\ \;\ \;\ \;\ \;\ \;\ \rightarrow \;\ \;\ \;\ 0, \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ cos(u)$

    Add up the signed products diagonally and alternate signs.

    $\displaystyle -u^{2}cos(u)+2usin(u)+2cos(u)+C$

    Resub:

    $\displaystyle \boxed{\frac{1}{2}\left[-(e^{2x})^{2}cos(e^{2x})+2e^{2x}sin(e^{2x})+2cos(e^ {2x})\right]}$

    Someone will come by saying, "this way is easier". Oh well. I wanted to show you tabular integration in the event you have not seen it. The more one knows the better.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    incidentally, a substitution works nicely with the first one also

    $\displaystyle \int \sin \sqrt{x}~dx$

    Let $\displaystyle u = \sqrt{x}$, that is, $\displaystyle u^2 = x$

    $\displaystyle \Rightarrow 2u~du = dx$

    So our integral becomes

    $\displaystyle 2 \int u \sin u~du$

    which is easy to do by parts. you can also use the tabular method as galactus showed you, but it would be overkill here, i think
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  7. #7
    Senior Member DeMath's Avatar
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    $\displaystyle \int {\sin \sqrt x dx = \int {\sqrt x \frac{{\sin \sqrt x }}{{\sqrt x }}} } dx = - 2\int {\sqrt x d\left( {\cos \sqrt x } \right)} =$

    $\displaystyle = - 2\sqrt x \cos \sqrt x + 2\int {\cos \sqrt x d\left( {\sqrt x } \right)} = - 2\sqrt x \cos \sqrt x + 2\sin \sqrt x + C.$
    Last edited by DeMath; Jan 23rd 2009 at 04:53 PM.
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