Integration problem

  • Thread starter EV33
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  • #1
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Integral of cos√x



We are supposed to use substitution and integration by parts but I really don't know where to even start.



No matter what I substitute for U I will be left without a du.
 

Answers and Replies

  • #2
gabbagabbahey
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what are your limits of integration?...what does your integral become when you use the substitution u=sqrt(x)? What is du?
 
  • #3
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There are no limits it is indefinite. If you let U= √x then du= 1/2√x and that is my problem because I am left with with integral of cos(u) and no du.
 
  • #4
gabbagabbahey
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doesn't [itex]du=\frac{1}{2\sqrt{x}}dx[/itex] and doesn't that mean that [itex]dx=2 \sqrt{x} du= 2udu[/itex]?
 
  • #5
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Yes it does. So does that mean when I integrate I get u^2 sinu and from there I just need to back substitute?
 
  • #6
gabbagabbahey
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Well, that means that

[itex]\int cos(\sqrt{x})dx= \int 2ucos(u)du[/itex]

now you'll need to integrate by-parts....try using f(u)=2u and g'(u)=cos(u)du
 
  • #7
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oh ok. Thank you for the help.
 

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