Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 44

We can rewrite this integral by expanding the sin(2x) into 2sin(x)cos(x).
int (cos(x) + sin(x))/(2sin(x)cos(x)) dx
which equals
1/2 int cos(x)/(sin(x)cos(x)) dx + 1/2 int sin(x)/(sin(x)cos(x)) dx
in the first integral we can cancel the cos(x) and in the second integral we can cancel the sin(x).
1/2 int (dx)/sin(x) + 1/2 int (dx)/cos(x)
1/sin(x) = csc(x) and 1/cos(x) = sec(x)
we can look up in a trig integral table that the integral of csc and sec are -ln(csc(x) + cot(x)) and ln(sec(x) + tan(x)) respectively.
so this whole integral is
1/2 (-ln(csc(x) + cot(x)) + ln(sec(x) + tan(x)))