How do you evaluate the integral #(secx tanx) / (sec^2(x) - secx)# dx?

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Answer 1 · 2018-05-10T11:52:38

#I=ln|1-cosx|+c#

We know that,
#color(red)((1)tanx=sinx/cosx and secx= 1/cosx#
Here,

#I=int(secxtanx)/(sec^2x-secx)dx#

#=int(cancelsecxtanx)/(cancelsecx(secx-1))dx#

#=inttanx/(secx-1)dx#

#=int(sinx/cosx)/(1/cosx-1)dx...tocolor(red)(Apply(1)#

#I=intsinx/(1-cosx)dx#

Let,

#1-cosx=u=>sinxdx=du#

#I=int1/udu#

#=ln|u|+c, where, u=1-cosx#

#I=ln|1-cosx|+c#