How do you find the integral of #int sin x * tan x dx#?

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Answer 1 · 2018-03-08T08:08:23

The answer is #=ln(|tanx+secx|)-sinx+C#

We need

#tanx=sinx/cosx#

#intsecxdx=ln(tanx+secx)+C#

Therefore,

#intsinxtanxdx=intsecxsin^2xdx=intsecx(1-cos^2x)dx#

#=int(secx-cosx)dx#

#=intsecxdx-intcosxdx#

#=ln(|tanx+secx|)-sinx+C#