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Math Help - integral with hyperbolic fn

  1. #1
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    integral with hyperbolic fn

    Hello,

    please if you can have a look through my workings.

    integr[(e^(4x))cosh5x]dx = (1/5)[(e^(4x))sinh5x] - (4/25)integr[(e^(4x))cosh5x] = (1/5)[(e^(4x))sinh5x] - (4/25)(e^(4x))cosh5x + (16/25)integr[(e^(4x))cosh5x)] => (9/25)integr[(e^(4x))cosh5x]dx = [(1/25)(e^(4x)](5sinh5x - 4cosh5x) = [(1/25)[(e^(9x))-9(e^(-x))] => integr[(e^(4x))cosh5x]dx = (1/9)(e^(9x))-(e^(-x))] . My book though says it equals
    (1/18)(e^(9x))-(1/2)(e^(-x))]

    thanks for any help
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  2. #2
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    Quote Originally Posted by pepsi View Post
    Hello,

    please if you can have a look through my workings.

    integr[(e^(4x))cosh5x]dx = (1/5)[(e^(4x))sinh5x] - (4/25)integr[(e^(4x))cosh5x] = (1/5)[(e^(4x))sinh5x] - (4/25)(e^(4x))cosh5x + (16/25)integr[(e^(4x))cosh5x)] => (9/25)integr[(e^(4x))cosh5x]dx = [(1/25)(e^(4x)](5sinh5x - 4cosh5x) = [(1/25)[(e^(9x))-9(e^(-x))] => integr[(e^(4x))cosh5x]dx = (1/9)(e^(9x))-(e^(-x))] . My book though says it equals
    (1/18)(e^(9x))-(1/2)(e^(-x))]

    thanks for any help
    Personally I'd just substitute the definition of cosh (5x) and apply a well known index law. The integrand becomes:

    e^{4x} \left( \frac{e^{5x} + e^{-5x}}{2}\right) = \frac{1}{2} \left( e^{9x} + e^{-x} \right).

    Now integrate.

    Much simpler, much less opportunity for carless mistakes. Life's too short for what you've done.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by pepsi View Post
    Hello,

    please if you can have a look through my workings.

    integr[(e^(4x))cosh5x]dx = (1/5)[(e^(4x))sinh5x] - (4/25)integr[(e^(4x))cosh5x] = (1/5)[(e^(4x))sinh5x] - (4/25)(e^(4x))cosh5x + (16/25)integr[(e^(4x))cosh5x)] => (9/25)integr[(e^(4x))cosh5x]dx = [(1/25)(e^(4x)](5sinh5x - 4cosh5x) = [(1/25)[(e^(9x))-9(e^(-x))] => integr[(e^(4x))cosh5x]dx = (1/9)(e^(9x))-(e^(-x))] . My book though says it equals
    (1/18)(e^(9x))-(1/2)(e^(-x))]

    thanks for any help
    Wow! Take a look...

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