Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 32

inttan^2(x)sec(x)dx
use the identity:tan^2(x)=sec^2(x)-1
inttan^2(x)sec(x)dx=int(sec^2(x)-1)sec(x)dx
=int(sec^3(x)-sec(x))dx
Now apply the Integral Reduction:intsec^n(x)dx=(sec^(n-1)(x)sin(x))/(n-1)+(n-2)/(n-1)intsec^(n-2)(x)dx
intsec^3(x)dx=(sec^2(x)sin(x))/2+1/2intsec(x)dx
Use the common integral:intsec(x)dx=ln(tan(x)+sec(x))
:.inttan^2(x)sec(x)dx=(sec^2(x)sin(x))/2+1/2intsec(x)dx-intsec(x)dx
=(sec^2(x)sin(x))/2-1/2intsec(x)dx
=(sec^2(x)sin(x))/2-1/2ln(tan(x)+sec(x))
add a constant C to the solution,
=(sec^2(x)sin(x))/2-1/2ln(tan(x)+sec(x))+C