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

Feb 11, 2018

$\int \setminus \sinh \frac{x}{1 + \cosh \left(x\right)} \setminus \mathrm{dx} = \ln \left(1 + \cosh \left(x\right)\right) + C$

#### Explanation:

We begin by introducing a u-substitution with $u = 1 + \cosh \left(x\right)$. The derivative of $u$ is then $\sinh \left(x\right)$, so we divide through by $\sinh \left(x\right)$ to integrate with respect to $u$:
$\int \setminus \sinh \frac{x}{1 + \cosh \left(x\right)} \setminus \mathrm{dx} = \int \setminus \frac{\cancel{\sinh \left(x\right)}}{\cancel{\sinh \left(x\right)} \cdot u} \setminus \mathrm{du} = \int \setminus \frac{1}{u} \setminus \mathrm{du}$

This integral is the common integral:
$\int \setminus \frac{1}{t} \setminus \mathrm{dt} = \ln | t | + C$

This makes our integral:
$\ln | u | + C$

We can resubstitute to get:
$\ln \left(1 + \cosh \left(x\right)\right) + C$, 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.