Evaluate $$I=\int \frac{\sec x \:dx}{\sin (2x+\theta)+\sin \theta}$$
I used $$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$$ we get
$$I=\frac{1}{2} \times \int \frac{\sec x \:dx}{\sin (x+\theta) \cos x}=\frac{1}{2} \times \int \frac{\sec^2 x \:dx}{\sin (x+\theta)} $$
Now i used $\tan x=y$ we get
$$2I=\int \frac{\sqrt{1+y^2}dy}{y \cos \theta+\sin \theta}$$
But how can we proceed from here?