Answer
$\frac{1}{7}\sec^7 x-\frac{2}{5}\sec^5 x+\frac{1}{3}\sec^3 x+C$
Work Step by Step
$\int\tan^5 x\sec^3 x\ dx$
Since the power of tangent is odd, save a factor of $\sec x\tan x$ and express the remaining factors in terms of $\sec x$:
$=\int\tan^4 x\sec^2 x\sec x\tan x\ dx$
$=\int(\tan^2 x)^2\sec^2 x\sec x\tan x\ dx$
$=\int(\sec^2 x -1)^2\sec^2 x\sec x\tan x\ dx$
Let $u=\sec x$. Then $du=\sec x\tan x\ dx$.
$=\int(u^2 -1)^2 u^2\ du$
$=\int(1-2u^2+u^4)u^2\ du$
$=\int(u^6-2u^4+u^2)\ du$
$=\frac{1}{7}u^7-\frac{2}{5}u^5+\frac{1}{3}u^3+C$
$=\boxed{\frac{1}{7}\sec^7 x-\frac{2}{5}\sec^5 x+\frac{1}{3}\sec^3 x+C}$
