What is the Integral of #((sec x)^3/tan x) dx#?

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Answer 1 · 2018-06-13T19:41:59

#I=ln|cscx-cotx|+secx+C#

Here,

#I=int(secx)^3/tanx dx#

#=int(secx(color(red)(sec^2x)))/tanx dx#

#=int(secx(color(red)(1+tan^2x)))/tanx dx#

#=int(secx+secxtan^2x)/tanx dx#

#=intsecx/tanxdx+int(secxtan^2x)/tanx dx#

#=int(1/cosx)/(sinx/cosx) dx+intsecxtanx dx#

#=intcscxdx+intsecxtanxdx#

#=ln|cscx-cotx|+secx+C#

Answer 2 · 2018-06-13T19:49:14

# 1/2ln|(cosx-1)/(cosx+1)|+1/cosx+C, or, #,

# lnsqrt|((cosx-1)/(cosx+1))|+secx+C, or, #,

# ln|tan(x/2)|+secx+C#.

Suppose that, #I=intsec^3x/tanxdx#.

#:. I=int1/cos^3x*cosx/sinxdx=int1/(cos^2xsinx)dx#.

#=intsinx/(cos^2x*sin^2x)dx#,

#=intsinx/{cos^2x(1-cos^2x)}dx#,

Letting #cosx=t," so that, "-sinxdx=dt#, we have,

#I=int(-1)/(t^2(1-t^2))dt=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)|-(t^(-2+1)/(-2+1)}#,

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

Since, #t=cosx#, we have,

# I=1/2ln|(cosx-1)/(cosx+1)|+1/cosx#,

#=lnsqrt|((cosx-1)/(cosx+1))|+secx#,

# rArr I=ln|tan(x/2)|+secx+C#.