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Math Help - integral x^2 Sinh(x)dx

  1. #1
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    integral x^2 Sinh(x)dx

    Hi again. Trying to solve this by parts. Question is evaluate

    \int\limits_0^{\ln 2} {x^2 } Sinh(x)\_dx

    This is what I get and how I got there.

    = x^2 Cosh(x) - \int {2xSinh(x)} \_dx \\
      = x^2 Cosh(x) - 2\int {xCosh(x)} \_dx \\
      = x^2 Cosh(x) - 2[xSinh(x) - \int {Sinh(x)\_dx} ] \\
      = x^2 Cosh(x) - 2[xSinh(x) - Cosh(x)] \\
      = (\ln 2)^2 Cosh(x) - 2[(\ln 2)Sinh(\ln 2) - Cosh(\ln 2)] \\
      = \frac{5}{4}(\ln 2)^2  - \frac{3}{2}(\ln 2) + \frac{5}{2} \\

    However the answer on the sheet is as below. What am I doing wrong?
    <br />
 = \frac{5}{4}(\ln 2)^2  - \frac{3}{2}(\ln 2) + \frac{1}{2}
    Last edited by Craka; June 4th 2008 at 07:55 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Craka View Post
    Hi again. Trying to solve this by parts. Question is evaluate

    \int\limits_0^{\ln 2} {x^2 } Sinh(x)\_dx

    This is what I get and how I got there.

    = x^2 Sinh(x) - \int {2xSinh(x)} \_dx \\
     = x^2 Sinh(x) - 2\int {xCosh(x)} \_dx \\
     = x^2 Sinh(x) - 2[xSinh(x) - \int {Sinh(x)\_dx} ] \\
     = x^2 Sinh(x) - 2[xSinh(x) - Cosh(x)] \\
     = (\ln 2)^2 Sinh(x) - 2[(\ln 2)Sinh(\ln 2) - Cosh(\ln 2)] \\
     = \frac{5}{4}(\ln 2)^2 - \frac{3}{2}(\ln 2) + \frac{5}{2} \\

    However the answer on the sheet is as below. What am I doing wrong?
    <br />
= \frac{5}{4}(\ln 2)^2 - \frac{3}{2}(\ln 2) + \frac{1}{2}
    I think your first mistake would be

    I assumed you let u=x^2 and dv=\sinh(x)dx

    then du=2xdx and v=\color{red}{\cosh(x)}

    unless I am mistaken
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  3. #3
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    Thanks for you reply, but sorry I don't follow. I did let u=x^2 dv= Sinh(x)
    du= 2x dx and v= Cosh(x)
    Are you saying should be v= -Cosh(x) ?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Craka View Post
    Thanks for you reply, but sorry I don't follow. I did let u=x^2 dv= Sinh(x)
    du= 2x dx and v= Cosh(x)
    Are you saying should be v= -Cosh(x) ?
    Oh, I am sorry that was a typo, yes you are correct in that v=cosh(x)

    But isnt the formula for parts \int{udv}=uv-\int{vdu}

    So dont you have

    x^2\sinh(x) which is equal to udv?
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  5. #5
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    Sorry I made a type in posting my working I have now corrected it, but question still remains on what I have done wrong.
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Craka View Post
    Sorry I made a type in posting my working I have now corrected it, but question still remains on what I have done wrong.
    Quote Originally Posted by Craka View Post
    Hi again. Trying to solve this by parts. Question is evaluate

    \int\limits_0^{\ln 2} {x^2 } Sinh(x)\_dx

    This is what I get and how I got there.

    = x^2 Cosh(x) - \int {2xSinh(x)} \_dx \\
      = x^2 Cosh(x) - 2\int {xCosh(x)} \_dx \\
      = x^2 Cosh(x) - 2[xSinh(x) - \int {Sinh(x)\_dx} ] \\
      = x^2 Cosh(x) - 2[xSinh(x) - Cosh(x)] \\
      = (\ln 2)^2 Cosh(x) - 2[(\ln 2)Sinh(\ln 2) - Cosh(\ln 2)] \\
      = \frac{5}{4}(\ln 2)^2  - \frac{3}{2}(\ln 2) + \frac{5}{2} \\

    However the answer on the sheet is as below. What am I doing wrong?
    <br />
 = \frac{5}{4}(\ln 2)^2  - \frac{3}{2}(\ln 2) + \frac{1}{2}
    I'm gonna take a shot at this..I haven't dealt with hyperbolic trig in a while..

    I believe you have it right, up to here:

      = x^2 Cosh(x) - 2[xSinh(x) - Cosh(x)] \\

    Apply the FTC, we get:

      =\left[ (\ln(2))^2 \cosh(\ln(2)) - 2[\ln(2)\sinh(\ln(2)) - \cosh(\ln(2))\right] {\color{red}-\left[-2(-\cosh(0))\right]}<br />
    =\frac{5}{4}(\ln(2))^2-\frac{3}{2}\ln(2)+\frac{5}{2}-2
    =\color{red}\boxed{\frac{5}{4}(\ln(2))^2-\frac{3}{2}\ln(2)+\frac{1}{2}}

    Hope this makes sense!
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  7. #7
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    Where/why did the -Cosh(0) come from?
    Are you applying the FTC thereom of f(b)-f(a) and if so would you not substitute 0 through all of the expression to evaluate?
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Craka View Post
    Where did =\left[ (\ln(2))^2 \cosh(\ln(2)) - 2[\ln(2)\sinh(\ln(2)) - \cosh(\ln(2))\right] {\color{red}-\left[-2(-\cosh(0))\right]}

    come from. If doing that wouldn't you sub zero in for the whol expression? I take it as you are doing f(b)-f(a) ??
    Yes, you would substitute into the whole thing, but note that \sinh(0)=0. Thus, all those terms drop out except for the last one.

    Yes, I was doing f(b)-f(a).
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  9. #9
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    Sorry that is exactly what you are doing. Just realised of my stupidity.
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