My post here<\/a> might be a helpful warm-up towards the technique below.<\/p>\n For \u222bsec\u00b3xdx we use integration by parts:<\/p>\n \u222buv’ = uv – \u222bvu’<\/p>\n With sec\u00b3x we imagine v’=sec\u00b2xdx, u=secx. Note that if v’=sec\u00b2xdx, v=tanx. Also, if u=secx, u’=secxtanxdx:<\/p>\n \u222bsec\u00b3xdx = secxtanx – \u222btanx(secxtanx)dx = secxtanx – \u222btan\u00b2xsecxdx<\/p>\n tan\u00b2x = sec\u00b2x – 1 ⇒ \u222btan\u00b2xsecxdx = \u222b(sec\u00b3x – secx)dx<\/p>\n \u222bsec\u00b3xdx = secxtanx – \u222b(sec\u00b3x – secx)dx = secxtanx – \u222bsec\u00b3xdx + \u222bsecxdx<\/p>\n So far, we have<\/p>\n \u222bsec\u00b3xdx = secxtanx – \u222bsec\u00b3xdx + \u222bsecxdx<\/p>\n Adding \u222bsec\u00b3xdx to both sides gives<\/p>\n 2\u222bsec\u00b3xdx = secxtanx + \u222bsecxdx<\/p>\n From the tables we know that \u222bsecxdx = ln|secx + tanx|:<\/p>\n 2\u222bsec\u00b3xdx = secxtanx + ln|secx + tanx|<\/p>\n Dividing both sides by 2, we arrive at<\/p>\n \u222bsec\u00b3xdx = (secxtanx + ln|secx + tanx|)\/2 + C<\/p>\n Source:<\/p>\n Larson, Roland E. and Robert P. Hostetler. Calculus<\/u>, third ed. Toronto: D C Heath and Company, 1989.<\/p>\n