Question:

Let $I = \int \frac{\sin x + \cos x}{9 + 16 \sin 2x} dx$<br>Put $\sin x - \cos x = t \Rightarrow \sin 2x = 1 - t^2$ and $(\sin x + \cos x) dx = dt$<br>$\therefore I = \int \frac{dt}{25 - 16t^2} = \frac{1}{2 \times 5 \times 4} \ln \left| \frac{5 + 4t}{5 - 4t} \right| + C = \frac{1}{40} \ln \left| \frac{5 + 4\sin x - 4\cos x}{5 - 4\sin x + 4\cos x} \right| + C$