How do you integrate #tanx/(1+cosx)#?

1 Answer
Jul 4, 2016

#inttanx/(1+cosx)dx=intsinx/(cosx(1+cosx))dx#

let

#u=cosx#
#du=-sinxdx#

partial fractions

#-int(du)/(u(u+1))=-int(1/u-1/(u+1))du#

#-int(1/u-1/(u+1))du=int(du)/(u+1)-int(du)/u#

#-int(du)/u=-ln|u|#

For #int(du)/(u+1)#

set
#w=u+1#
#dw=du#

#int(dw)/w = ln|w|#

Plug back in for u and x

#ln|w|= ln|u+1|#

So far we have

#ln|u+1|-ln|u|#

Remember #u=cosx#

#ln|cosx+1|-ln|cosx|#

And #+ C#

#inttanx/(1+cosx)dx=ln|cosx+1|-ln|cosx|+C#