$\int \frac{\mathrm{d}x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}(a,b>0)$ is equal to

 

Putting $t=\tan x$ we get

$\begin{array}{l}\int \frac{\mathrm{d}x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}=\int \frac{{\sec }^{2}x\mathrm{d}x}{a^2+b^2{\tan }^{2}x}=\int \frac{\mathrm{d}t}{a^2+b^2{t}^2}\\ =\frac{1}{b^2}\int \frac{\mathrm{d}t}{t^2+\left(a^2/b^2\right)}=\frac{1}{b^2}\cdot \frac{b}{a}{\tan }^{-1}\left(\frac{b}{a}t\right)+C\\ =\frac{1}{ab}{\tan }^{-1}\left(\frac{b}{a}\tan x\right)+C.\end{array}$