How was this integration problem $\int \frac{\sec^2 x}{\tan x}\:dx=\log |\tan x | +C$ solved?

Question

$$\int \frac{\sec^2 x}{\tan x}\:dx=\log |\tan x | +C$$

When I started working on the problem, I used $1 + \tan^2x = \sec^2x$, but that didn't get me anywhere. How has the problem been solved?

Answer 1

Observe that $$ \frac{d}{dx}\tan x=(\tan x)'=\sec^2 x $$ giving, with the change of variable $u=\tan x$, $du= \sec^2 x \:dx$ $$ \int \frac{\sec^2 x}{\tan x}\:dx=\int \frac{du}{u}=\log |u|+C=\log |\tan x | +C $$ for any constant $C$.

Answer 2

Hint: $$\frac{d}{dx}(\tan x)=1+\tan^2x=\sec^2x$$