# Integrate cosh x/(3 sinh x +4)

• Jun 3rd 2010, 07:48 AM
MechEng
Integrate cosh x/(3 sinh x +4)
Given

$
\int{\frac{cosh(x)}{3 sinh(x)+4}}
$

I come up with

$let u=sinh(x), du=cosh(x)$

$\int{\frac{du}{3u+4}}$

So

$\frac{1}{3}\int{\frac{du}{u}}+4$

And

$\frac{1}{3} sinh(x)+4x+C$

So, how many rules did I violate or otherwise offend with this one?
• Jun 3rd 2010, 08:05 AM
Unknown008
I never worked with sinh nor cosh but I can tell you this:

$\frac{1}{x+1} \neq \frac{1}{x} + 1$

What you need to do from

$\int{\frac{du}{3u+4}}$

is to use ln.

If you differentiate the denominator, you get 3, so:

$\int{\frac{du}{3u+4}} = \frac{1}{3} (ln(3u+4)) + c$

From there, you can substitute back. :)
• Jun 3rd 2010, 08:08 AM
General
Personally, I will substitute $u=3sinh(x)+4$ ..
• Jun 3rd 2010, 08:31 AM
MechEng
Ok...

Then it should look something like...

$
\int{\frac{cosh(x)}{3 sinh(x)+4}}
$

let

$
u=3sinh(x)+4
$

$
du=3cosh(x)
$

So...

$
\int{\frac{\frac{1}{3}du}{u}}
$

And...

$
\frac{1}{3}\int{\frac{du}{u}}
$

And...

$\frac{1}{3}u+C$

So...

$\frac{1}{3}(3sinh(x)+4)+C$

And finally...

$
sinh(x) + \frac{4}{3} +C
$
• Jun 3rd 2010, 08:44 AM
Unknown008
You seem to have forgotten your integrations...

$\int \frac{1}{x} dx = ln(x) + c$

or;

$\int \frac{f'(x)}{f(x)} dx = ln(f(x)) + c$
• Jun 3rd 2010, 08:52 AM
General
Quote:

Originally Posted by Unknown008
You seem to have forgotten your integrations...

$\int \frac{1}{x} dx = ln{\color{red} |}x{\color{red} |} + c$

or;

$\int \frac{f'(x)}{f(x)} dx = ln{\color{red} |}f(x){\color{red} |} + c$

..
• Jun 3rd 2010, 08:58 AM
Unknown008
Quote:

Originally Posted by General
..

Yes, but in my education system, they do not put the absolute symbols, and we are not given marks for them in the exams, so I tend to forget those...(Itwasntme)
• Jun 3rd 2010, 09:16 AM
MechEng
Doh...

$
\frac{1}{3}ln|3sinh(x)+4|+C
$

I just reviewed logarithmic differentiation and integration. I can't believe I missed that.

I don't recall seeing the absolute value being ephasized in my book either. Ah, nevermind, there it is.

$
\int{\frac{dx}{x}}= ln|x|+C
$
• Jun 3rd 2010, 09:21 AM
General
Quote:

Originally Posted by MechEng
Doh...
$