Question:

$$\int \frac{dx}{4 \sin^2 x + 12 \sin x \cos x + 9 \cos^2 x}$$<br>Divide numerator and denominator by $\cos^2 x$<br>$$\therefore I = \int \frac{\sec^2 x dx}{4 \tan^2 x + 12 \tan x + 9} = \int \frac{\sec^2 x dx}{(2 \tan x + 3)^2}$$<br>Let $2 \tan x + 3 = t \Rightarrow 2 \sec^2 x dx = dt$<br>$$I = \frac{1}{2} \int \frac{dt}{t^2} = -\frac{1}{2t} + C = -\frac{1}{2(2 \tan x + 3)} + C$$