Question

$int sec x dx$

Explanation:

Step1: Multiply by $\frac{\sec x+\tan x}{\sec x+\tan x}$

$\int\sec x\cdot\frac{\sec x+\tan x}{\sec x+\tan x}dx=\int\frac{\sec^{2}x+\sec x\tan x}{\sec x+\tan x}dx$

Step2: Use substitution

Let $u = \sec x+\tan x$, then $du=(\sec x\tan x+\sec^{2}x)dx$. So the integral becomes $\int\frac{du}{u}$.

Step3: Integrate $\frac{1}{u}$

We know that $\int\frac{du}{u}=\ln|u|+C$.

Step4: Substitute back $u$

Substituting $u = \sec x+\tan x$ back, we get $\ln|\sec x+\tan x|+C$.

Answer:

$\ln|\sec x+\tan x|+C$