Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 29

intsec^5(x)tan^3(x)dx
Let's rewrite the integral as:
intsec^5(x)tan^3(x)dx=intsec^5(x)tan^2(x)tan(x)dx
Now using the trigonometric identity:tan^2(x)=sec^2(x)-1
=intsec^5(x)(sec^2(x)-1)tan(x)dx
=intsec^4(x)(sec^2(x)-1)sec(x)tan(x)dx
Now apply the integral substitution:u=sec(x)
du=sec(x)tan(x)dx
=intu^4(u^2-1)du
=int(u^6-u^4)du
apply the sum rule,
=intu^6du-intu^4du
=(u^(6+1)/(6+1))-(u^(4+1)/(4+1))
=u^7/7-u^5/5
substitute back u=sec(x) and add a constant C to the solution,
=(sec^7(x))/7-(sec^5(x))/5+C