Question:

Let $I = \int \frac{dx}{3 \sin x + 4 \cos x} = \int \frac{dx}{3 \left\{ \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right\} + 4 \left\{ \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right\}} = \int \frac{\sec^2 \frac{x}{2} dx}{4 + 6 \tan \frac{x}{2} - 4 \tan^2 \frac{x}{2}}$<br>Put $\tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^2 \frac{x}{2} dx = dt$<br>So $I = \int \frac{2 dt}{4 + 6t - 4t^2} = \frac{1}{2} \int \frac{dt}{1 - \left(t^2 - \frac{3}{2}t\right)} = \frac{1}{2} \int \frac{dt}{\frac{25}{16} - \left(t - \frac{3}{4}\right)^2}$