Let J=∫sec5(x)dx=∫(sec3(x))(sec2(x))dx.
Let u=sec3(x), du=3sec2(x)(sec(x)tan(x))dx=3sec3(x)tan(x)dx
let dv=sec2(x)dx, v=tan(x).
J=sec3(x)tan(x)-∫3sec3(x)tan2(x)dx.
tan2(x)=sec2(x)-1, so:
J=sec3(x)tan(x)-3∫sec3(x)(sec2(x)-1)dx,
J=sec3(x)tan(x)-3∫sec5(x)dx+3∫sec3(x)dx,
J=sec3(x)tan(x)-3J+3∫sec3(x)dx,
4J=sec3(x)tan(x)+3∫sec3(x)dx.
Let K=∫sec3(x)dx=∫(sec(x))(sec2(x))dx.
Let u=sec(x), du=sec(x)tan(x)dx, v=tan(x) as before.
K=sec(x)tan(x)-∫sec(x)tan2(x)dx=sec(x)tan(x)-K+∫sec(x)dx,
2K=sec(x)tan(x)+∫sec(x)dx,
2K=sec(x)tan(x)+∫sec(x)(sec(x)+tan(x))dx/(sec(x)+tan(x)),
2K=sec(x)tan(x)+∫(sec2(x)+sec(x)tan(x))dx/(sec(x)+tan(x)),
2K=sec(x)tan(x)+ln|sec(x)+tan(x)|, K=½(sec(x)tan(x)+ln|sec(x)+tan(x)|).
J=∫sec5(x)dx=¼(sec3(x)tan(x)+3∫sec3(x)dx)=
¼(sec3(x)tan(x)+(3/2)(sec(x)tan(x)+ln|sec(x)+tan(x)|)+C, where C is integration constant.