How do you find integral of #((secxtanx)/(secx-1))dx#?

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Answer 1 · 2015-04-13T16:52:59

#int(sec(x).tan(x))/(sec(x)-1)dx#

We know that :

#sec(x) = 1/cos(x)#

#tan(x) = sin(x)/cos(x)#

So, we have :

#=int1/cos(x).sin(x)/cos(x)*1/(1/cos(x)-1)dx#

#= int(sin(x))/(cos(x)-cos^2(x))dx#

Let's #t=cos(x)#

So #dt = -sin(x)#

#= int1/(t-t^2)(-dt)#

Factorize
#- int1/(t(1-t))dx #

With decomposition in simple elements, we have :

#alpha/t+beta/(1-t)#

#= (alpha(1-t))/(t(1-t))+(betat)/(t(1-t))#

So :

#alpha-alphat+betat#
#alpha+t(-alpha+beta)#

We have :

#alpha = 1#
#-alpha+beta=0#

#alpha=beta=1#

We get :

#-int1/tdt+int1/(1-t)dt#

#-[log(t)]+[log(1-t)]#

#=-[log(cos(x))]+[log(1-cos(x))]#

#=2log(sin(x/2))-log(cos(x))+C#

Answer 2 · 2015-04-13T17:11:29

Another way of doing this is to consider that #d(secx) = secxtanx*dx#

That is, the derivative of #secx# is #secxtanx#

#=> int(secxtanx)/(secx - 1)dx = int d(secx)/(secx - 1)#

This is the same as letting #u = secx#

We then have,

#int(du)/(u - 1) = ln(u - 1) + C = ln(secx - 1) + C#