Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 23

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
To evaluate the given integral problem int tan^5(x/2) dx , we may apply u-substitution by letting: u = x/2 then du =1/2 dx or 2du= dx .
The integral becomes:
int tan^5(x/2) dx =int tan^5(u)* 2 du
Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int tan^5(u)* 2 du =2 int tan^5(u)du
Apply integration formula for tangent function: int tan^n(x)dx = (tan^(n-1)(x))/(n-1)- int tan^(n-2)(x)dx .
2 int tan^5(u)du= 2 *[(tan^(5-1)(u))/(5-1)- int tan^(5-2)(u)du]
= 2*[(tan^(4)(u))/(4)- int tan^(3)(u)du]
Apply another set integration formula for tangent function on int tan^(3)(u)du .
int tan^(3)(u)du = (tan^(3-1)(u))/(3-1)- int tan^(3-2)(u)du
= (tan^(2)(u))/(2)- int tan^(1)(u)du
=(tan^(2)(u))/(2)-ln (sec(u))+C
Applying int tan^(3)(u)du =(tan^(2)(u))/(2)-ln (sec(u))+C , we get:
2 int tan^5(u)du=2*[(tan^(4)(u))/(4)- int tan^(3)(u)du]
=2*[(tan^(4)(u))/(4)- [(tan^(2)(u))/(2)-ln (sec(u))]]+C
=2*[(tan^(4)(u))/(4)-(tan^(2)(u))/(2)+ln (sec(u))]+C
=(tan^(4)(u))/2-tan^(2)(u)+2ln (sec(u))+C
Plug-in u = x/2 on (tan^(4)(u))/2-tan^(2)(u)+2ln (sec(u))+C , we get the indefinite integral as:
int tan^5(x/2) dx=(tan^(4)(x/2))/2-tan^(2)(x/2)+2ln (sec(x/2))+C