Answer
\[\frac{\sec x\;\tan x}{2}-\frac{1}{2}\ln|\sec x+\tan x|+C\]
Work Step by Step
Let $\;\;$ $I=\int\tan^2 x\:\sec x\,dx$ ____(1)
$I=\int\tan x\:(\sec x\:\tan x)dx$
[$\sec^2 x-\tan^2 x=1$]
$I=\int\sqrt{\sec^2 x-1}\:(\sec x\:\tan x)dx$
Substitute $t=\sec x$ ___(2)
$\Rightarrow dt=\sec x\:\tan xdx$
$I=\int\sqrt{t^2-1} \;dt$
$\left[\int\sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln|x+\sqrt{x^2-a^2}|\right]$
$I=\frac{t}{2}\sqrt{t^2-1}-\frac{1}{2}\ln|t+\sqrt{t^2-1}|+C$
From (2)
$I=\frac{\sec x}{2}\sqrt{\sec^2 x-1}-\frac{1}{2}\ln|\sec x+\sqrt{\sec^2 x-1}|+C$
$I=\frac{\sec x\:\tan x}{2}-\frac{1}{2}\ln|\sec x+\tan x|+C$
Hence $I=\frac{\sec x\:\tan x}{2}-\frac{1}{2}\ln|\sec x+\tan x|+C$.
