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Integration substitution help

  1. Feb 17, 2005 #1
    [tex]\int \frac {cos(\sqrt{x})}{\sqrt{x}}dx =?[/tex]

    Here's what I did:
    [tex]= \int x^{-0.5}cosx^{0.5}dx [/tex]
    subsitute:
    [tex]u= cos(\sqrt{x})[/tex]
    [tex]du=-sin(\sqrt{x})(0.5x^{-0.5})dx[/tex]
    [tex]-\frac {1}{0.5sin(\sqrt{x})}\int u du[/tex]
    [tex]-\frac{2}{sin(\sqrt{x})} 0.5cos^2(\sqrt{x})[/tex]
    [tex]-\frac{1}{sin(\sqrt{x})}cos^2(\sqrt{x})[/tex]

    I know I did this wrong. Any suggestions?
     
    Last edited: Feb 17, 2005
  2. jcsd
  3. Feb 17, 2005 #2

    Galileo

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    Be careful. You cannot remove [itex]\frac{1}{\sin(\sqrt(x))}[/itex] from the integral, because it depends on x! (u depends on x too).

    You just made a bad choice for substitution. No biggy, just try a different one. Not too many obvious option left anymore..
     
  4. Feb 17, 2005 #3

    HallsofIvy

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    Well, first of all you can't just take that "[itex]sin(\sqrt{x})[/itex]" out of the integral- it's a function of x!

    DON'T substitute of the whole cos x0.5, just for x0.5.

    Let u= x0.5 so that du= 0.5x-0.5dx and 2du= x-0.5.

    Now, its easy!
     
  5. Feb 17, 2005 #4
    Try substituting u=x^1/2...

    oops...beat to the punch...
     
  6. Feb 17, 2005 #5
    is the answer [tex] 2sin(\sqrt{x})+c?[/tex]
     
  7. Feb 17, 2005 #6
    yep...u can always check by deriving the answer and see if you get the original function.
     
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