Calculus of a Single Variable, Chapter 5, 5.2, Section 5.2, Problem 36

int(2-tan(theta/4))d theta=
Use additivity of integral: int(f(x)pm g(x))dx=int f(x)dx pm int g(x)dx. int2d theta-int tan(theta/4)d theta=
Since the first integral is easy int 2d theta=2theta+C we will concentrate on the second integral. To solve it we will make substitution: u=theta/4, du=(d theta)/4=>d theta=4du
int tan(theta/4)d theta=4int tan u du=-4ln|cos u|+C
Return the substitution.
-4ln|cos(theta/4)|+C
Now we subtract the two integrals to obtain the final result.
2theta-(-4ln|cos(theta/4)|)+C=2theta+4ln|cos(theta/4)|+C