How do you find the antiderivative of #ln(cosx)tanxdx#?

Question

2391 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-04T15:15:06

#-(ln(cosx))^2/2+C#

We wish to find:

#intln(cosx)tanxdx#

Use substitution:

If #u=ln(cosx)#, then #du=(-sinx)/cosxdx=-tanxdx#.

Thus,

#intln(cosx)tanxdx=-intln(cosx)(-tanx)dx=-intudu#

This becomes

#-u^2/2+C=-(ln(cosx))^2/2+C#