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

Question

9507 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-13T23:08:38

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

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

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:

$=ln(absv)+C$

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

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

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