Question:

Let $I = \int \frac{dx}{3 + 4 \sin 2x} = \int \frac{dx}{3 + \frac{4 \cdot 2 \tan x}{1 + \tan^2 x}} = \int \frac{(1 + \tan^2 x) dx}{3 + 3 \tan^2 x + 8 \tan x}$<br>Put $\tan x = t \Rightarrow \sec^2 x dx = dt$<br>$\therefore I = \int \frac{dt}{3t^2 + 8t + 3} = \frac{1}{3} \int \frac{dt}{\left(t + \frac{4}{3}\right)^2 - \frac{7}{9}}$<br>$= \frac{1}{3} \cdot \frac{3}{2\sqrt{7}} \ln \left| \frac{t + \frac{4}{3} - \frac{\sqrt{7}}{3}}{t + \frac{4}{3} + \frac{\sqrt{7}}{3}} \right| + C = \frac{1}{2\sqrt{7}} \ln \left| \frac{\tan x + \frac{4}{3} - \frac{\sqrt{7}}{3}}{\tan x + \frac{4}{3} + \frac{\sqrt{7}}{3}} \right| + C$