Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 46

Determine the integral $\displaystyle \int \frac{\cos x + \sin x }{\sin 2 x} dx$

$\begin{equation} \begin{aligned} \int \frac{\cos x + \sin x }{\sin 2 x} dx &= \int \frac{\cos x + \sin x }{2 \sin x \cos x} dx \qquad \text{Apply Trigonometric Identity } \sin 2x = 2 \sin x \cos x\\ \\ \int \frac{\cos x + \sin x }{\sin 2 x} dx &= \int\left( \frac{\cancel{\cos x}}{2 \sin x \cancel{\cos x}} + \frac{\cancel{\sin x}}{2 \cancel{\sin x} \cos x} \right)dx\\ \\ \int \frac{\cos x + \sin x }{\sin 2 x} dx &= \int \left( \frac{1}{2\sin x} + \frac{1}{2\cos x} \right) dx\\ \\ \int \frac{\cos x + \sin x }{\sin 2 x} dx &= \int \left( \frac{1}{2} \csc x + \frac{1}{2} \sec x \right) dx\\ \\ \int \frac{\cos x + \sin x }{\sin 2 x} dx &= \frac{1}{2} \left[ -\ln (\csc x + \cot x) + \ln ( \sec x + \tan x) \right] + c \qquad \text{ or } \qquad \int \frac{\cos x + \sin x }{\sin 2 x} dx = \frac{1}{2} \left[ \ln(\sec x + \tan x) - \ln (\csc x + \cot x) \right] + c \end{aligned} \end{equation}$