Results 1 to 9 of 9

Thread: integral x^2 Sinh(x)dx

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    175

    integral x^2 Sinh(x)dx

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

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

    This is what I get and how I got there.

    $\displaystyle = x^2 Cosh(x) - \int {2xSinh(x)} \_dx \\ $
    $\displaystyle = x^2 Cosh(x) - 2\int {xCosh(x)} \_dx \\ $
    $\displaystyle = x^2 Cosh(x) - 2[xSinh(x) - \int {Sinh(x)\_dx} ] \\ $
    $\displaystyle = x^2 Cosh(x) - 2[xSinh(x) - Cosh(x)] \\ $
    $\displaystyle = (\ln 2)^2 Cosh(x) - 2[(\ln 2)Sinh(\ln 2) - Cosh(\ln 2)] \\$
    $\displaystyle = \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?
    $\displaystyle
    = \frac{5}{4}(\ln 2)^2 - \frac{3}{2}(\ln 2) + \frac{1}{2}$
    Last edited by Craka; Jun 4th 2008 at 07:55 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Craka View Post
    Hi again. Trying to solve this by parts. Question is evaluate

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

    This is what I get and how I got there.

    $\displaystyle = x^2 Sinh(x) - \int {2xSinh(x)} \_dx \\ $
    $\displaystyle = x^2 Sinh(x) - 2\int {xCosh(x)} \_dx \\ $
    $\displaystyle = x^2 Sinh(x) - 2[xSinh(x) - \int {Sinh(x)\_dx} ] \\ $
    $\displaystyle = x^2 Sinh(x) - 2[xSinh(x) - Cosh(x)] \\ $
    $\displaystyle = (\ln 2)^2 Sinh(x) - 2[(\ln 2)Sinh(\ln 2) - Cosh(\ln 2)] \\$
    $\displaystyle = \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?
    $\displaystyle
    = \frac{5}{4}(\ln 2)^2 - \frac{3}{2}(\ln 2) + \frac{1}{2}$
    I think your first mistake would be

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

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

    unless I am mistaken
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2008
    Posts
    175
    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) ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    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 $\displaystyle \int{udv}=uv-\int{vdu}$

    So dont you have

    $\displaystyle x^2\sinh(x)$ which is equal to $\displaystyle udv$?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2008
    Posts
    175
    Sorry I made a type in posting my working I have now corrected it, but question still remains on what I have done wrong.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    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

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

    This is what I get and how I got there.

    $\displaystyle = x^2 Cosh(x) - \int {2xSinh(x)} \_dx \\ $
    $\displaystyle = x^2 Cosh(x) - 2\int {xCosh(x)} \_dx \\ $
    $\displaystyle = x^2 Cosh(x) - 2[xSinh(x) - \int {Sinh(x)\_dx} ] \\ $
    $\displaystyle = x^2 Cosh(x) - 2[xSinh(x) - Cosh(x)] \\ $
    $\displaystyle = (\ln 2)^2 Cosh(x) - 2[(\ln 2)Sinh(\ln 2) - Cosh(\ln 2)] \\$
    $\displaystyle = \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?
    $\displaystyle
    = \frac{5}{4}(\ln 2)^2 - \frac{3}{2}(\ln 2) + \frac{1}{2}$
    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:

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

    Apply the FTC, we get:

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

    Hope this makes sense!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jun 2008
    Posts
    175
    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?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    Quote Originally Posted by Craka View Post
    Where did $\displaystyle =\left[ (\ln(2))^2 \cosh(\ln(2)) - 2[\ln(2)\sinh(\ln(2)) - \cosh(\ln(2))\right] {\color{red}-\left[-2(-\cosh(0))\right]}$

    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 $\displaystyle \sinh(0)=0$. Thus, all those terms drop out except for the last one.

    Yes, I was doing f(b)-f(a).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Jun 2008
    Posts
    175
    Sorry that is exactly what you are doing. Just realised of my stupidity.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 26th 2010, 12:30 PM
  2. Sinh and Skn
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Aug 29th 2009, 06:24 AM
  3. Replies: 3
    Last Post: Apr 27th 2009, 02:18 AM
  4. integral with log and sinh
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Aug 8th 2008, 05:47 PM
  5. sinh
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jun 19th 2008, 06:15 AM

Search Tags


/mathhelpforum @mathhelpforum