The tutor shows how to integrate sec³x.

My post here might be a helpful warm-up towards the technique below.

For ∫sec³xdx we use integration by parts:

∫uv’ = uv – ∫vu’

With sec³x we imagine v’=sec²xdx, u=secx. Note that if v’=sec²xdx, v=tanx. Also, if u=secx, u’=secxtanxdx:

∫sec³xdx = secxtanx – ∫tanx(secxtanx)dx = secxtanx – ∫tan²xsecxdx

tan²x = sec²x – 1 ⇒ ∫tan²xsecxdx = ∫(sec³x – secx)dx

∫sec³xdx = secxtanx – ∫(sec³x – secx)dx = secxtanx – ∫sec³xdx + ∫secxdx

So far, we have

∫sec³xdx = secxtanx – ∫sec³xdx + ∫secxdx

Adding ∫sec³xdx to both sides gives

2∫sec³xdx = secxtanx + ∫secxdx

From the tables we know that ∫secxdx = ln|secx + tanx|:

2∫sec³xdx = secxtanx + ln|secx + tanx|

Dividing both sides by 2, we arrive at

∫sec³xdx = (secxtanx + ln|secx + tanx|)/2 + C

Source:

Larson, Roland E. and Robert P. Hostetler. Calculus, third ed. Toronto: D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.