Expert: Ahmed Salami - 12/7/2005

Question
intergrate 1/3+4coshx

Answer
Hi spax,
Forgive the time taken.
For our purpose, $ represents the integral sign.
By definition,
cosh x = (e^x + e^-x)/2
$dx/(3 + 4cosh x) becomes
$dx/(3 + 2e^x + 2e^-x)
We use the substitution u = e^x
du/dx = e^x = u
dx = du/u
also, e^-x = 1/e^x
We can therefore simplify the integral to
$du/ u(3 + 2u + 2/u)
= $du/(2u^2 + 3u + 2)
= 1/2 $du/ [u^2 + (3/2)u + 1]
We have to make use of the standard integral form
$dz/(z^2 + A^2) = (1/A)arctan (z/A)
We rearrange u^2 + (3/2)u + 1 by completing the
square to give
u^2 + (3/2)u + (3/4)^2 + 1 - (3/4)^2
= (u + 3/4)^2 + (sqrt7 /4)^2
1/2 $du/ [u^2 + (3/2)u + 1] becomes
1/2 $du/ (u + 3/4)^2 + (sqrt7 /4)^2
We can now make z = (u + 3/4)
dz = du
and A = (sqrt7 /4)
Our final result then becomes
1/2 .1/(sqrt7 /4). arctan [(u + 3/4)/(sqrt7 /4)]
= 2(sqrt7)/7 arctan [(4u + 3)(sqrt7)/7]
Returning back to u = e^x
$dx/(3 + 4cosh x) = 2(sqrt7)/7 arctan [(4e^x + 3)(sqrt7)/7]
I have omitted some simplifications hoping you'll get by, but you
can always get back to me. I hope i have helped you.
Regards.