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Math Help - Another problem on Integration by parts

  1. #1
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    Another problem on Integration by parts

    integrate e^x coshx dx
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  2. #2
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    Quote Originally Posted by confusedgirl View Post
    integrate e^x coshx dx
    if only you could write this as the integral of e^x[something involving e^x and e^(-x)]


    This bit was only good to show 0=0 so no piont keeping it
    Last edited by jiboom; February 8th 2010 at 08:51 AM.
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  3. #3
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    As jiboom suggested, the simplest thing to do is to use the definition of cosh(x) in terms of e^x. What are you using as the definition of cosh(x)?
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    (e^x + e^-x)/2
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    Quote Originally Posted by jiboom View Post
    if only you could write this as the integral of e^x[something involving e^x and e^(-x)]

    or,if you have to do it by parts, just use parts twice. If we call the integral I., when you do parts twice you will end up with

    I=something+kI whrer something and k are found from doing parts

    which we can arrange as

    (1-k)I=something

    and so I can be found.
    if i integrate by parts 2 times
    the result will be I = 0/2 which is equal to zero
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  6. #6
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    Quote Originally Posted by confusedgirl View Post
    integrate e^x coshx dx
    Cosh(x)=\frac{e^x+e^{-x}}{2}

    e^xCosh(x)=\frac{e^xe^x+e^xe^{-x}}{2}=\frac{e^{2x}+e^0}{2}

    =\frac{e^{2x}+1}{2}

    \frac{1}{2}\int{e^{2x}}dx+\int{\frac{1}{2}}dx=\fra  c{1}{2}\int{e^xe^x}dx+\frac{x}{2}+C

    To integrate by parts

    Let \int{e^xe^x}dx=I

    u=e^x\ \Rightarrow\ du=e^xdx

    dv=e^xdx\ \Rightarrow\ v=e^x

    uv-\int{v}du=e^xe^x-\int{e^xe^x}dx

    I=e^{2x}-I

    2I=e^{2x}

    I=\frac{e^{2x}}{2}+C

    Therefore

    \int{e^xCosh(x)}dx=\frac{1}{4}e^{2x}+\frac{1}{2}x+  C
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  7. #7
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    Quote Originally Posted by confusedgirl View Post
    if i integrate by parts 2 times
    the result will be I = 0/2 which is equal to zero


    i only added that as an edit as you wanted parts. I have never done that sort of integral by parts, i would use my first (and useful comment).
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