How do you integrate #int e^-xtan(e^-x)dx#?

1 Answer
Dec 13, 2016

Answer:

#inte^-xtan(e^-x)dx=ln(abscos(e^-x))+C#

Explanation:

#inte^-xtan(e^-x)dx#

First, let #u=e^-x#. Differentiating this shows that #du=-e^-xdx#.

#=-inttan(e^-x)(-e^-x)dx#

#=-inttan(u)du#

This is a standard integral, but we can show how to integrate it by rewriting tangent as sine divided by cosine:

#=-intsin(u)/cos(u)du#

Now, let #v=cos(u)#. This implies that #dv=-sin(u)du#.

#=int(-sin(u))/cos(u)du#

#=int(dv)/v#

This is also a standard (and very important) integral:

#=ln(absv)+C#

#=ln(abscos(u))+C#

#=ln(abscos(e^-x))+C#