What's the integral of int sinx * tanxdx?

1 Answer
Nov 11, 2015

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