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Homework Help: Integration by Parts

  1. Jul 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the solution to:

    y' = x.y.cos(x^2)

    2. Relevant equations

    Integration by Parts method.

    3. The attempt at a solution

    Step 1
    (dy/dx).(1/y) = x.cos(x2)

    (1/y) dy = x.cos(x2) dx

    Step 2
    Integrate both sides.

    ln|y| = integratal of [ x.cos(x2) dx ]

    Step 3
    Using integration by parts...

    u = cos(x2) => du = -2x.sin(2^x)
    dv = x => v = 1.dx

    Step 4
    Subbing back in...

    ln|y| = u.v - integral of v.du = cos(x2) - integral of [-2x.sin(x2)]

    ln(y) = cos(x2) + 2*integral of [x.sin(x2)]

    Step 5
    Using integration by parts a second time...

    u = sin(x2) => du = 2x.cos(x2)
    dv = x => v = 1.dx

    Step 6
    Subbing back in...

    ln|y| = cos(x2) + 2 ( sin(x2) - 2*integral of x.cos(x2).dx )

    Step 7

    I stop my attempt there because it just seems to eventually I get to a point where it becomes a function of itself? (ie. the x.cos(x2)

    What do I do next? or have I gone wrong somewhere?


    Ok so I noticed that we get "integral of [x.cos(x2)]" back in our formula, and up above in Step 2 I declared it to be = ln|y|

    So I sub ln|y| into the formula and get...

    ln|y| = cos(x2) + 2sin(x2) - 4*ln|y|
    5*ln|y| = cos(x2) + 2sin(x2)
    ln|y| = (1/5)*(cos(x2) + 2sin(x2))

    y = e^(above line)

    Ta da? lemmy know if this correct please!
    Last edited: Jul 21, 2011
  2. jcsd
  3. Jul 21, 2011 #2
    integral of [ x.cos(x2) dx ] doesn't need integration by parts right? derivative of x square is 2 times x.
  4. Jul 21, 2011 #3

    Are you high? :P
    What has the derivative of x^2 got to do with the integral of x*cos(2^x) ?
  5. Jul 21, 2011 #4
    Why are you even using integration by parts? A much simpler approach is to substitute t=x^2 on your right hand integration.
  6. Jul 21, 2011 #5
    HEY! I WASN'T HIGH! I was trying to help!
  7. Jul 21, 2011 #6
    As far as I'm aware the easiest way to integrate the product of two functions is to integrate it by parts.

    Could you please explain your method? ie. Once I substitute t=x^2, what then?
  8. Jul 21, 2011 #7
    If t=x^2, dt=2xdx. Your integral xcos(x2)dx can then be expressed as (1/2)cos(t)dt, which you can integrate easily.
  9. Jul 21, 2011 #8

    How does x become (1/2) ?
  10. Jul 21, 2011 #9
    You haven't done integrations by substitution yet? Maybe you should wait for your class to get to that point...(Actually if you are doing differential equations, you should be aware of this)

    Anyway: xcos(x2)dx = (1/2)cos(x2)(2xdx) = (1/2)cos(t)dt
  11. Jul 21, 2011 #10


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    This is pretty standard when using substitution to do integration. Generally the method of
    substitution is covered before integration by parts.

    You are trying to find [itex]\displaystyle \int x\cos(x^2)\,dx\,.[/itex]

    As Oster suggested, use substitution. Pi-Bond gave more details: use t = x2, then dt = 2x dx → x dx = (1/2) dt . This is where the 1/2 comes from.

    x doesn't become 1/2. Pi-Bond moved the x to be with 2 x dx, then had to include 1/2 to make up for the extra factor of 2. It's just a different way to get the same result.
  12. Jul 21, 2011 #11

    Thanks guys, understand it all now. Legends.
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