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Math Help - Integration by parts

  1. #1
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    Integration by parts

    I am half way stuck with this integral, any help would be appreciated.

    Integral (p^5)ln(p) dp

    u = ln p, du = 1/p
    dv = p^5, v = p^6/6

    Integral (p^5)ln(p) dp = p^6/6 ln(p) - Integral p^6/6p dp

    Im stuck at the integral, substitution doesn't seem to work if I let w = p^6, dw = 6p^5..
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  2. #2
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    Quote Originally Posted by ff4930 View Post
    Integral p^6/6p dp
    \int\frac{p^6}{6p}\,dp=\frac16\int p^5\,dp

    No need substitution, just apply the power rule.
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  3. #3
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    Hello again, I stumbled with another difficult integral and need help. I didn't want to waste bandwidth so I added to this current thread.

    I have integral e^2x sin3x dx

    u = sin3x, du = 3cos3x
    dv = e^2x, v = 1/2e^2x

    integral e^2x sin3x dx = 1/2 e^2x sin3x - 3/2 integral e^2x cos3x dx

    for integral e^2x cos3x dx
    w = cos3x, dw = -3sin3x
    dv = e^2x, v = 1/2e^2x

    integral e^2x cos3x dx = 1/2e^2x cos3x + 3/2 integral e^2x sin3x dx

    so combined is..

    integral e^2x sin3x dx = 1/2 e^2x sin3x - 3/4 e^2x cos3x - 9/4 integral e^2x sin3x dx

    since I have integral e^2x sin3x dx on both sides can I add integral e^2x sin3x dx making it

    2 integral e^2x sin3x dx = 1/2 e^2x sin3x - 3/4 e^2x cos3x - 9/4

    then divide both sides by 2

    getting...

    integral e^2x sin3x dx = [1/2 e^2x sin3x - 3/4 e^2x cos3x - 9/4]/2 + C?

    Thanks for all your help.
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  4. #4
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    First, set y=3x, then apply integration by parts.
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  5. #5
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    Sorry, I don't quite understand what you mean by set y = 3x.
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  6. #6
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    Quote Originally Posted by ff4930 View Post
    integral e^2x sin3x dx
    y=3x\implies dy=3\,dx, the integral becomes

    \int e^{2x}\sin3x\,dx=\frac13\int e^{\frac23y}\sin y\,dy

    It's to simply a little bit the integrand . Now apply integration by parts.
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  7. #7
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    It did simplified but I'm not good with variable fraction exponents.

    so 1/3 integral e^(2/3y) sin y dy

    u = sin y, du = cos y
    dv = e^(2/3y) v = ???

    unless Im not seeing something Im suppose to.
    Sorry if Im not getting it right away.
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  8. #8
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    Hello, ff4930!

    A small arithmetic error in there . . . I'll do it from scratch.


    We have: . \int e^{2x}\sin3x\,dx

    By parts: . \begin{array}{ccccccc}u & = & \sin3x & \quad & dv & = & 3\cos3x\,dx \\<br />
du & = & 3\cos3x\,dx & \quad & v & = &\frac{1}{2}e^{2x}\end{array}

    Then: . I \;=\;\frac{1}{2}e^{2x}\sin3x - \frac{3}{2}\int e^{2x}\cos3x\,dx


    By parts again:\ . \begin{array}{ccccccc}u & = & \cos3x & \quad & dv & = & e^{2x}dx \\ du & = & -3\sin3x\,dx & \quad & v & = & \frac{1}{2}e^{2x} \end{array}

    And we have: . I \;=\;\frac{1}{2}e^{2x}\sin3x - \frac{3}{2}\left[\frac{1}{2}e^{2x}\cos3x + \frac{3}{2}\int e^{2x}\sin3x\,dx\right]

    . . which simplifies to: . I \;=\;\frac{1}{2}e^{2x}\sin3x - \frac{3}{4}e^{2x}\cos3x - \frac{9}{4}\underbrace{\int e^{2x}\sin3x\,dx}_{\text{This is }I}


    So we have: . I \;=\;\frac{1}{2}e^{2x}\sin3x - \frac{3}{4}e^{2x}\cos3x - \frac{9}{4}I + C

    . . Then: . \frac{13}{4}I \;=\;\frac{1}{2}e^{2x}\sin3x - \frac{3}{4}e^{2x}\cos3x + C \;=\;\frac{1}{4}e^{2x}\left(2\sin3x - 3\cos3x\right) + C

    Multiply by \frac{4}{13}\!:\;\;I \;=\;\frac{1}{13}\left(2\sin3x - 3\cos3x\right) + C


    . . Therefore: . \int e^{2x}\sin3x\,dx \;=\;\frac{1}{13}\left(2\sin3x - 3\cos3x\right) + C
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  9. #9
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    Hi!, thanks for replying.

    I'm confuse on how you get
    13/4 on the left side, also did you add 9/4 I on both sides? so shouldn't it be 2I 13/4?


    \frac{13}{4}I \;=\;\frac{1}{2}e^{2x}\sin3x - \frac{3}{4}e^{2x}\cos3x + C \;=\;\frac{1}{4}e^{2x}\left(2\sin3x - 3\cos3x\right) + C<br />
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  10. #10
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    Quote Originally Posted by Soroban View Post
    . . Therefore: . \int e^{2x}\sin3x\,dx \;=\;\frac{1}{13}\left(2\sin3x - 3\cos3x\right) + C
    Actually, it should be \int e^{2x}\sin3x\,dx=\frac1{13}e^{2x}(2\sin3x-3\cos3x)+k

    In general

    \int e^{ax}\sin(bx)\,dx=\frac{e^{ax}(a\sin(bx)-b\cos(bx))}{a^2+b^2}+k

    You (to ff4930) can derive this formula, of course, applyin' integration by parts.
    Last edited by Krizalid; September 4th 2007 at 10:46 AM. Reason: Forgot the constant
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