#intsec(2x)/tan(x)dx=#?

1 Answer
maganbhai P.
May 2, 2018

#I=1/2ln|tan^2x/(tan^2x-1)|+c#

Explanation:

Here,

#I=int(sec2x)/tanxdx#

We know that,

#=cos2x=(1-tan^2x)/(1+tan^2x)=>sec2x=(1+tan^2x)/(1-tan^2x#

So,

#I=int(1+tan^2x)/(tanx(1-tan^2x))dx=intsec^2x/(tanx(1- tan^2x))dx#

Let, #tanx=t=>sec^2xdx=dt#

#:.I=int1/(t(1-t^2))dt#

#=int[1/t+t/(1-t^2)]dt#

#=int[1/t-t/(t^2-1)]dt#

#=int1/tdt-1/2int(2t)/(t^2-1)dt#

#=ln|t|-1/2int(d/(dt)(t^2-1))/(t^2-1)dt#

#=ln|t|-1/2ln|t^2-1|+c#

#=1/2[2ln|t|-ln|t^2-1|]+c#

#=1/2[ln|t^2|-ln|t^2-1|]+c#

#=1/2ln|t^2/(t^2-1)|+c,where, t=tanx#

#=>I=1/2ln|tan^2x/(tan^2x-1)|+c#

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