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Homework Help: U-substitution integration help

  1. Nov 29, 2006 #1
    1. The problem statement, all variables and given/known data

    [tex]\int cos(1/2)x = 2sin(x/2) + c[/tex] ==> Correct Answer

    This is an answer to one of my homework problems, but I don't understand why the indefinite integtral of cos(1/2)x is that. I've learned u-substitution and the power rule, but using those methods, I got:


    2. The attempt at a solution


    [tex] u = .5x, du = .5dx \int cos(u)du = sin(u)+c = sin(.5x)+c[/tex] (wrong answer?)
     
  2. jcsd
  3. Nov 29, 2006 #2

    radou

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    You made a mistake while substituting. [tex]du = \frac{1}{2}dx[/tex] implies [tex]dx = 2 du[/tex].
     
  4. Nov 29, 2006 #3
    [tex]\int cos(1/2)x = 2sin(x/2) + c[/tex]

    Yeah I'm confused. The integrand is equal to

    2/sqrt(2)*x so the integral is equal to (1/sqrt(2))x^2 + C
     
    Last edited: Nov 29, 2006
  5. Nov 29, 2006 #4
    Let [tex] u = \frac{1}{2}x [/tex]. Then [tex] du = \frac{1}{2} dx [/tex].


    So you have: [tex]2 \int \cos u \; du [/tex]
     
    Last edited: Nov 29, 2006
  6. Nov 29, 2006 #5

    radou

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    What is there to be confused about? And where did you get these 'sqrt-s' from? As said before, you only have to substitute dx = 2 du and u = 1/2 x into the original integral, in order to obtain [tex]\int 2 \cos u du[/tex], which I assume you'll know how to handle.
     
  7. Nov 29, 2006 #6
    I think I got the right answer:

    [tex]\int cos(1/2)x dx = 2 \int cos(.5x)(.5) + c = 2sin(.5x) + c[/tex]

    But I don't understand how you got that.
     
  8. Nov 29, 2006 #7

    radou

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    OK, again: [tex]\int \cos(\frac{1}{2}x)[/tex] = after substituting = [tex]\int 2 \cos u du = 2 \sin u + C = 2 \sin (\frac{1}{2}x) + C[/tex] .
     
  9. Nov 29, 2006 #8
    Thank you , radou. I understand it now.

    Nevermind what I said in post #6; I just realize that nealh149 was talking about another problem.
     
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