Let $I=\int\tan^4xdx$ $\int(\sec^2x-1)^2dx$ $=\int(\sec^4x-2\sec^2x+1)dx$ $=\int\sec^4xdx-2\int\sec^2xdx+\int1 dx$ $=\int\sec^4xdx-2\tan x+x+C_1$ $\Rightarrow I= I_1-2\tan x+x+C_1$ (say).....(i)where $I_1=\int\sec^4dx$Now, $I_1=\int\sec^4dx$ $=\int\sec^2x \cdot \sec^xdx$ $=\int(1+\tan^2x)\sec^2xdx$Put $\tan x = t \Rightarrow \sec^2 x dx = dt$$\therefore I_1=\int(1+t^2)dt$ $=t+\frac{t^3}{3}+C_2$ $=\tan x+\frac{1}{3}\tan^3x+C_2$...(ii)From (i) and (ii), we have$I= \tan x+\frac{1}{3}\tan^3x+C_2-2\tan x+x+C_1$$= \frac{1}{3}\tan^3x - \tan x+x+C$