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`