What's the integral of #int sinx * tanxdx#?

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Answer 1 · 2015-11-11T00:37:32

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

let's #t = sin(x)#

#dt = cos(x)#

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

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

#=-[t]+[arctanh(t)]+C#

Substitute back for #t=sin(x)#

#-[sin(x)]+[arctanh(sin(x))]+C#