Answer
$$\int4\tan^3xdx=2\tan^2x-4\ln|\sec x|+C$$
Work Step by Step
$$A=\int4\tan^3xdx=\int4\tan x\tan^2xdx$$ $$A=\int4\tan x(\sec^2x-1)dx$$ $$A=4\Big(\int\tan x\sec^2xdx-\int\tan xdx\Big)$$ $$A=4\Big(\int\tan xd(\tan x)-\ln|\sec x|\Big)$$
We set $u=\tan x$.
$$A=4\Big(\int udu-\ln|\sec x|\Big)$$ $$A=4\Big(\frac{u^2}{2}-\ln|\sec x|\Big)+C$$ $$A=2u^2-4\ln|\sec x|+C$$ $$A=2\tan^2x-4\ln|\sec x|+C$$
