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Simple Integration by Parts

  1. Jul 23, 2008 #1
    1. The problem statement, all variables and given/known data
    #1.
    Use Integration by parts to evaluate the integral
    [tex]\int 2x \ln(2x) dx[/tex]

    #2.
    Use Integration by parts to evaluate the integral
    [tex]\int (\ln(3x))^{2} dx[/tex]

    #3.
    Use Integration by parts to evaluate the integral
    [tex]\int x e^{4x} dx[/tex]

    #4.
    Evaluate the indefinite integral.
    [tex]\int \sin(3x) \sin(11x) dx[/tex]


    3. The attempt at a solution
    #1.
    [tex]\int fg'\ = fg - \int f'g\[/tex]
    f=2x f'=2
    g'=ln(2x) g=1/2x
    [tex]\int fg'\ = \frac {2x}{2x}\ - \int \frac{2}{2x}\\[/tex]
    = 1 - ln|x| + C
    I think I went wrong finding the anti-derivative of ln(2x). Would g(x) be 1/2 * 1/2x instead?

    #2.
    [tex]\int (\ln(3x))^{2} dx[/tex]
    Make both f(x) and g'(x) = ln(3x)?
    I think I'm getting this question wrong because I don't know the anti-derivative of ln(3x).

    #3.
    [tex]\int x e^{4x} dx[/tex]
    f=x f'=1
    g'=e^4x g = 1/4 e^4x
    = [1/4 * x * e^4x - 1/4 * e^4x] + C

    #4.
    [tex]\int \sin(3x) \sin(11x) dx[/tex]
    [tex]1/2 \int \cos(3x-11x) - \cos(3x+11x) dx[/tex]
    [tex]1/2 \int \cos(-8x) - \cos(14x) dx[/tex]
    [tex]1/2 [\sin(-8x) - \sin(14x)] dx[/tex]

    I'm not sure what I'm doing wrong here.
    Does sin(-8x) - sin(14x) = -sin(6x)?
     
  2. jcsd
  3. Jul 23, 2008 #2
    #1

    d/dx (x^2.ln(x)) = 2x.lnx + x

    integrate both things, and I find this way the easiest


    #2

    you would find anti-derivative for ln(3x) from the integration table. You probably will have that during your exam too

    #3

    I forgot the technique but I don't think your one is right, differentiate it
    the answer should be '1/16*(-1+4*x)*exp(4*x)' figure out how to reach there ...

    #4

    [tex]
    1/2 \int \cos(-8x) - \cos(14x) dx
    [tex]

    is not equal to your final answer. You forgot to take care of constants. and cos(-a) = cos(a)

    "Does sin(-8x) - sin(14x) = -sin(6x)? "
    x = 1
    does Does sin(-8*1) - sin(14*1) = -sin(6*1)??
     
    Last edited: Jul 23, 2008
  4. Jul 23, 2008 #3
    #1.
    Not sure how that way works. But can you confirm if the following are correct:
    f=2x f'=2
    g'=ln(2x) g=1/(2x)

    #2.
    The anti-derivative for ln(3x) = 1/3x?
    f=ln(3x) f'=(1/x)
    g'=ln(3x) g=?

    #3. SOLVED
    I did something wrong when first trying this question. I didn't integrate e^4x properly so i missed the 1/4.
    so [tex]\frac {(4x-1)*e^{4x}}{16}[/tex] was my final answer after simplifying from [tex]\frac {x}{4} e^{4x} - \frac {1}{16} e^{4x}[/tex].
    Thanks rootX.

    #4. SOLVED
    I totally forgot about taking care of the constants. Thank you rootX. Final answer is 1/2 * 1/8 [sin(8x)] - 1/2 * 1/14 [sin(14x)].
     
    Last edited: Jul 23, 2008
  5. Jul 23, 2008 #4
    no.. d/dx(ln(x)) = 1/x not the integral check your integration table

    Editing over.
     
  6. Jul 23, 2008 #5
    #5.
    Use integration by parts to evaluate the definite integral.
    [tex]
    \int t e^{-t} dt
    [/tex]

    Is the following correct?
    f = t f' = 1
    g' = [tex]e^{-t}[/tex] g = [tex]-e^{-t}[/tex]

    [tex]-t e^{-t} ]^{1}_{0} - [e^{-t}]^{1}_{0}[/tex]

    Solved Thanks Rocomath.
     
    Last edited: Jul 23, 2008
  7. Jul 23, 2008 #6
    Yes, keep going!

    Also, stick to "t", you used x for f.
     
  8. Jul 23, 2008 #7
    http://math2.org/math/integrals/more/ln.htm

    I don't think you need to know that.
     
  9. Jul 23, 2008 #8
    Cool, at first I didn't understand why it didn't matter if ln(2x) or ln(kx). Now I'm all cleared up! Thank you again rootX
     
  10. Jul 23, 2008 #9
    I thought you are messing derivatives with integration

    as int(ln(x)) = 1/x ...
     
  11. Jul 23, 2008 #10
    #2.
    [tex]\int (ln(3x))^{2} dx[/tex]

    I'm still stuck on this one.

    Do I use [tex]\int ln(3x) ln(3x)[/tex] and f=ln(3x) g'=ln(3x)

    or

    [tex]\int (ln(3x))^{2}[/tex] and f=ln(3x)^{2} g'=1
     
  12. Jul 23, 2008 #11
    try

    f = [ln(3x)]^2 and g'=x

    d/dx (x.[ln(3x)]^2 )= [ln(3x)]^2 + ..

    I am not familar with that your notations, I never tried to learn them
     
  13. Jul 23, 2008 #12
    I ended up with [tex]x(ln3x)^{2} - 2\int ln(3x)[/tex]

    u = [tex](ln(3x))^{2}[/tex] du = [tex]2 ln(3x) \frac {1}{x}[/tex]

    Not sure how to integrate ln(3x). I know integral of ln(x) is xln(x)-x+C. Still not sure what to do when it's ln(3x). I am now assuming any integration of ln(kx) is 1/x
     
    Last edited: Jul 23, 2008
  14. Jul 23, 2008 #13

    Defennder

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    Homework Helper

    [tex]\ln A + \ln B = \ln (AB)[/tex]
     
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