Answer
$\dfrac{\sec^5 x}{5}-\dfrac{\sec^3 x}{5}+c$
Work Step by Step
Consider the integral $\int \tan^3 x \sec^3 x dx$
The given integral can be re-written as:
$\int \tan^2 x \sec^2 x (\sec x \tan x) dx=\int (\sec^2 x -1) \sec^2 x (\sec x\tan x dx) $
Plug in $a= \sec x \implies da= \sec x \tan x dx$
We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$
This implies that $\int (a^4 -a^2) da =\dfrac{a^5}{5}-\dfrac{a^3}{3}+C=\dfrac{\sec^5 x}{5}-\dfrac{\sec^3 x}{5}+c$
