Question

integral hyperbolic function

Answer 1

\[\int\limits \frac{ 1 }{ sinhx*coshx }dx\]

Answer 2

hint:
we can rewrite your function as below:
\[\Large \frac{1}{{\sinh x\cosh x}} = \frac{{\cosh x}}{{\sinh x{{\left( {\cosh x} \right)}^2}}} = \frac{1}{{\tanh x}}\frac{1}{{{{\left( {\cosh x} \right)}^2}}}\]

Answer 3

furthermore, we have:
\[\Large d\left( {\tanh x} \right) = \frac{{dx}}{{{{\left( {\cosh x} \right)}^2}}}\]

Answer 4

ok would there be a way to do it just as simply without tanh?

Answer 5

you can try using the standard definitions of sinhx and coshx, namely:
\[\Large \sinh x = \frac{{{e^x} - {e^{ - x}}}}{2},\quad \cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}\]

Answer 6

alright thank you

Answer 7

thank you!