Answer
$$\int\frac{\sec^2 x}{\tan^2 x}dx=-\cot x+C$$
Work Step by Step
$$A=\int\frac{\sec^2 x}{\tan^2 x}dx$$
Let $u=\tan x$.
We would have $du=\sec^2 xdx$.
Substitute into $A$, we have $$A=\int\frac{1}{u^2}du$$ $$A=\int u^{-2}du$$ $$A=\frac{u^{-1}}{-1}+C$$ $$A=-\frac{1}{u}+C$$ $$A=-\frac{1}{\tan x}+C$$ $$A=-\cot x+C$$
