Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 484: 27

$\frac{1}{3}\sec^3 x- secx+C$

$\int\tan^3 x\sec x\ dx$ Since the power of tangent is odd, save a factor of $\sec x\tan x$ and express the remaining factors in terms of $\sec x$: $=\int\tan^2 x\sec x\tan x\ dx$ $=\int(\sec^2 x-1)\sec x\tan x\ dx$ Let $u=\sec x$. Then $du=\sec x\tan x\ dx$. $=\int(u^2-1)du$ $=\frac{1}{3}u^3-u+C$ $=\boxed{\frac{1}{3}\sec^3 x-sec(x)+C}$