Thread: integral with hyperbolic fn

1. 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

2. Originally Posted by pepsi
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:

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

Now integrate.

Much simpler, much less opportunity for carless mistakes. Life's too short for what you've done.

3. Originally Posted by pepsi
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...

http://www.mathhelpforum.com/math-he...-tutorial.html