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

Answer 1

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

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.

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Answer 2

Emma Aldridge

To evaluate the integral ( \int \frac{\sinh x}{1 + \cosh x} ):

Substitute ( u = \cosh x ).
Rewrite ( \sinh x ) in terms of ( u ) using the identity ( \sinh^2 x = \cosh^2 x - 1 ).
Rewrite the integral in terms of ( u ).
Use partial fraction decomposition to simplify the integrand.
Integrate the resulting expression with respect to ( u ).
Substitute back ( \cosh x ) for ( u ) in the final result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.