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Math Help - 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.

    \int sin \sqrt x~dx

    \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.

    \int sin \sqrt x~dx

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

    Thanks!
    For the first one, do you mean  \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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  4. #4
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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.

    \int sin \sqrt x~dx

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

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

     <br />
\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<br />

    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.

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

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

    Make the subs and we get:

    \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.

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

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

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

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

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

    Add up the signed products diagonally and alternate signs.

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

    Resub:

    \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

    \int \sin \sqrt{x}~dx

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

    \Rightarrow 2u~du = dx

    So our integral becomes

    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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    \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)}  =

    =  - 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; January 23rd 2009 at 04:53 PM.
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