How do you evaluate the integral #int sinhx/(1+coshx)#?

1 Answer
Feb 11, 2018

#int\ sinh(x)/(1+cosh(x))\ dx=ln(1+cosh(x))+C#

Explanation:

We begin by introducing a u-substitution with #u=1+cosh(x)#. The derivative of #u# is then #sinh(x)#, so we divide through by #sinh(x)# to integrate with respect to #u#:
#int\ sinh(x)/(1+cosh(x))\ dx=int\ cancel(sinh(x))/(cancel(sinh(x))*u)\ du=int\ 1/u\ du#

This integral is the common integral:
#int\ 1/t\ dt=ln|t|+C#

This makes our integral:
#ln|u|+C#

We can resubstitute to get:
#ln(1+cosh(x))+C#, which is our final answer.

We remove the absolute value from the logarithm because we note that #cosh# is positive on its domain so it's not necessary.