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Homework Help: Integration by parts

  1. Jun 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi there. I'm confused about this exercise. It asks me to solve the integral using integration by parts. And the integral is:

    [tex]\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx[/tex]

    3. The attempt at a solution
    What I did:

    [tex]u=3x[/tex]
    [tex]du=3dx[/tex]
    [tex]dv=cos(\displaystyle\frac{x}{2})[/tex]
    [tex]v=\sin(\displaystyle\frac{x}{2})[/tex]

    Then:

    [tex]\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=3x\sin(\displaystyle\frac{x}{2})-\displaystyle\int_{}^{}\sin(\displaystyle\frac{x}{2})3dx=3x\sin(\displaystyle\frac{x}{2})-3\cos(\displaystyle\frac{x}{2})[/tex]

    And derive gives me:

    [tex]\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=12\cos(\displaystyle\frac{x}{2})+6x\sin(\displaystyle\frac{x}{2})[/tex]

    So I think I'm doing something wrong, but I don't know what. And actually in my first attempt to a solution I got something like: [tex]\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=\displaystyle\frac{1}{3}[x\sin(\displaystyle\frac{x}{2})-\cos(\displaystyle\frac{x}{2})][/tex], I arrived to this result by first taking the 3 out of the integral, and passing it to the other side at the end.

    Any help will be thanked.

    Bye there.
     
  2. jcsd
  3. Jun 26, 2010 #2
    You have the right approach but you made at least 3 silly mistakes. Your expression for "v" is off by a factor of 2 and you are off by a minus sign when you integrate sine. The final step is really silly when you factor out a 3 and get 1/3.
     
  4. Jun 26, 2010 #3

    HallsofIvy

    User Avatar
    Science Advisor

    This is incorrect. Let u= x/2. Then du= (1/2)dx so dx= 2du.
    [tex]\int cos(x/2)dx= 2\int cos(u)du= 2 sin(u)+ C= 2sin(x/2)+ C[/tex]
    If [itex]dv= cos(x/2)[/itex], then [itex]v= 2sin(x/2)[/itex], not just sin(x/2).

     
  5. Jun 26, 2010 #4
    Thanks :)
     
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