How do you evaluate the integral #int sec^3x/tanx#?

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Answer 1 · 2017-01-19T15:06:24

#1/2ln|(cosx-1)/(cosx+1)|+secx+C, or, ln|tan(x/2)|+secx+C#.

Let #I=intsec^3x/tanxdx=int(1/cos^3x)(cosx/sinx)dx#

#=int1/(cos^2xsinx)dx=intsinx/(cos^2xsin^2x)dx#

#:. I=-int{(-sinx)/{cos^2x(1-cos^2x)}dx#

Substituting #cosx=t," so that, "-sinxdx=dt#, we get,

#I=int1/{t^2(t^2-1)}dt=int{t^2-(t^2-1)}/{t^2(t^2-1)}dt#

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

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

#1/2ln|(t-1)/(t+1)|+1/t#.

Since, #t=cosx#, we have,

#I=1/2ln|(cosx-1)/(cosx+1)|+secx+C#.

Enjoy Maths.!

N.B.:-#I# can further be simplified as #ln|tan(x/2)|+secx+C#.