Evaluate the indefinite integral integrate of arcsec(x)dx

mathsworkmusic | Certified Educator

You need to integrate by parts

int u (dv)/dx dx = uv - int v (du)/dx dx

Let v = x implies (dv)/dx = 1

And u = arcsecx implies (du)/dx = 1/(|x| sqrt(x^2 -1)) (see wikipedia page below)

Then int arcsecx = xarcsecx - int x/(|x|sqrt(x^2-1)) dx

Using integral of one over square root of a binomial and checking the cases x < -1 and x>1 (the domain of x)

implies \int arcsecx = xarcsecx - ln|x + sqrt(x^2 -1)| + C answer