# A formula for the integral of exponentials and hyperbolic trig functions

• Mar 7th 2011, 06:03 AM
Hamed
A formula for the integral of exponentials and hyperbolic trig functions
Hello,
Do we have any formula for
Integral of $\int{e^(ax)\cosh(x)dx}$
and also $\int{e^(ax)\sinh(x)dx}$
• Mar 7th 2011, 06:13 AM
Plato
Quote:

Originally Posted by Hamed
Do we have any formula for
Integral of e^(at)cosh(bx)dx?
and also e^(at)sinh(bx)dx

First, are you sure you have written what you meant?

Then because $\dfrac{d(\sinh(bx))}{dx}=b\cosh(bx)$ I don't see much to do.
• Mar 7th 2011, 06:15 AM
Prove It
Assuming that you meant $\displaystyle \int{e^{ax}\cosh{bx}\,dx}$ and $\displaystyle \int{e^{ax}\sinh{bx}\,dx}$, for each you need to use Integration by Parts twice.
• Mar 7th 2011, 06:17 AM
Plato
Quote:

Originally Posted by Prove It
Assuming that you meant $\displaystyle \int{e^{ax}\cosh{bx}\,dx}$ and $\displaystyle \int{e^{ax}\sinh{bx}\,dx}$, for each you need to use Integration by Parts twice.

Please note in the OP it is $\displaystyle \int{e^{at}\cosh{bx}\,dx}$ and [tex] not $\displaystyle \int{e^{ax}\cosh{bx}\,dx}$.
• Mar 7th 2011, 06:26 AM
Prove It
Quote:

Originally Posted by Plato
Please note in the OP it is $\displaystyle \int{e^{at}\cosh{bx}\,dx}$ and [tex] not $\displaystyle \int{e^{ax}\cosh{bx}\,dx}$.

Which is why I wrote "Assuming that you meant..."
• Mar 7th 2011, 06:27 AM
HallsofIvy
Yes, but as you said before, that is very simple. Assuming it is in fact, $e^{ax}sinh(bx)$, perhaps the simplest thing to do is to write it as
$e^{ax}\left(\frac{e^{bx}- e^{-bx}}{2}\right)= \frac{e^{(a+b)x}- e^{(a- b)x}}{2}$
and integrate that.
• Mar 7th 2011, 06:41 AM
Hamed
x not t
for e^(ax)cos(bx)dx= [(e^(ax))/(a^2+b^2)]*(acosbx+bsinbx)

Can we have sth for e^(ax)cosh(bx)dx?
and e^(ax)sinh(bx)dx

How can I use TEX?
• Mar 7th 2011, 06:44 AM
Ackbeet
Quote:

Originally Posted by Hamed
How can I use TEX?

See here for a LaTeX tutorial.
• Mar 7th 2011, 06:45 AM
Plato
Quote:

Originally Posted by Hamed
x not t
for e^(ax)cos(bx)dx= [(e^(ax))/(a^2+b^2)]*(acosbx+bsinbx)
Can we have sth for e^(ax)cosh(bx)dx?
and e^(ax)sinh(bx)dx
How can I use TEX?

In that case use post #6. It is the best way to do it.
There is a LaTeX tutorial here.
• Mar 7th 2011, 06:51 AM
Hamed
I need sth like that formula!
Quote:

Originally Posted by Plato
In that case use post #6. It is the best way to do it.
There is a LaTeX tutorial here.

• Mar 7th 2011, 07:04 AM
Plato
Quote:

Originally Posted by Hamed
I need sth like that formula!

You should be able to do this for yourself.
$\displaystyle\int {\frac{1}
{2}\left( {e^{\left( {a + b} \right)x} - e^{\left( {a - b} \right)x} } \right)dx} = \frac{1}
{2}\left( {\frac{{e^{\left( {a + b} \right)x} }}
{{a + b}} - \frac{{e^{\left( {a - b} \right)x} }}
{{a - b}}} \right) + C$
• Mar 7th 2011, 08:59 AM
Prove It
Quote:

Originally Posted by Hamed
I need sth like that formula!

I don't know what you mean by "sth', but if you need to write your indefinite integral in terms of sines and cosines, use integration by parts twice.

Otherwise, if you want a quick way to perform the integration, first convert the hyperbolic functions to their exponential forms.