Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 21

Indefinite integrals are written in the form of int f(x) dx = F(x) +C
where: f(x) as the integrand
F(x) as the anti-derivative function
C as the arbitrary constant known as constant of integration
For the given integral problem int sec^3(pix) dx , we may evaluate this using u-substitution.
Let: u = pix then du = pi dx or (du)/pi =dx .
The integral becomes:
int sec^3(pix) dx =int sec^3(u) * (du)/pi
Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int sec^3(u) * (du)/pi =1/piint sec^3(u) du
Apply integration formula for secant function:
int sec^n(x) dx = (sec^(n-1)(x)sin(x))/(n-1) + (n-2)/(n-1) int sec^(n-2)(x) dx
We get:
1/piint sec^3(u) du =1/pi [(sec^(3-1)(u)sin(u))/(3-1) + (3-2)/(3-1) int sec^(3-2)(u) du]
=1/pi [(sec^2(u)sin(u))/(2) + (1)/(2) int sec^(1)(u) du]
For the integral of int sec^(1)(u) du or int sec^(u) du , we may apply int sec(theta) d theta = ln(sec(theta)+tan(theta))+C .
Then,int sec^(u) du =ln(sec(u)+tan(u))+C
The complete indefinite integral will be:
1/piint sec^3(u) du =1/pi [(sec^2(u)sin(u))/(3-1) + (1)/(2) int sec^(1)(u) du]
=1/pi [(sec^2(u)sin(u))/(2) + (1)/(2)[ln(sec(u)+tan(u))]]+C
=1/pi [(sec^2(u)sin(u))/2+ln(sec(u)+tan(u))/2]+C
=(sec^2(u)sin(u))/(2pi)+ln(sec(u)+tan(u))/(2pi)+C
or tan(u)/(2picos(u)) + 1/(2pi)ln((1+sin(u))/cos(u))+C
Plug-in u = pix , we get the final indefinite integral as:
int sec^3(pix) dx=(sec^2(pix)sin(pix))/(2pi)+ln(sec(pix)+tan(pix))/(2pi)+C
or tan(pix)/(2picos(pix)) + 1/(2pi)ln((1+sin(pix))/cos(pix))+C