Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 28

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 problem int sec^2(x/2)tan(x/2) dx has a integrand in the form of a trigonometric function.
To evaluate this, we may apply u-substitution by letting u = tan(x/2) .
Then, the derivative of u is:
du = sec^2(x/2) *(1/2) dx
Rearrange this into 2 du= sec^2(x/2) dx .
Plug-in the values on the int sec^2(x/2)tan(x/2) dx , we get:
int sec^2(x/2)tan(x/2) dx =int u *2 du
Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int u *2 du =2int u du
Apply the Power Rule for integration:int (x^n) dx = x^(n+1)/ (n+1) +C .
2int u du =2* u^(1+1)/(1+1) +C
= 2 *u^2/2+C
= u^2 +C
Plug-in u = tan(x/2) on u^2 +C , we get the indefinite integral as:
int sec^2(x/2)tan(x/2) dx =(tan(x/2))^2 +C or tan^2(x/2) +C