Evaluating $\int\frac{dx}{(a\sin x+ b\cos x)^2}$, $a\neq 0.$

Question

Could you just show the hint to solve this integral, please?

Answer 1

Another way using the tangent half-angle subsitution $$\int\frac{dx}{(a\sin x+ b\cos x)^2}=2\int\frac{2 \left(t^2+1\right)}{\left(2 a t-b t^2+b\right)^2}\,dt=-\frac{2 t}{b \left(-2 a t+b t^2-b\right)}$$

Answer 2

Hint. One may write $$ \int\frac{dx}{(a\sin x+ b\cos x)^2}=\int\frac{1}{(a\tan x+b)^2}\frac{dx}{(\cos x)^2}=\int\frac{du}{(a\:u+b)^2} $$ with the change of variable $u=\tan x$.