Answer
$$\int\sec^4x\tan^2xdx=\frac{\tan^5x}{5}+\frac{\tan^3x}{3}+C$$
Work Step by Step
$$A=\int\sec^4x\tan^2xdx$$ $$A=\int(\sec^2x\tan^2x)(\sec^2x)dx$$ $$A=\int(\sec^2x\tan^2x)d(\tan x)$$
Take $u=\tan x$, which means $\tan^2x=u^2$
Also, since $\sec^2x=\tan^2x+1$, we would have $\sec^2x=u^2+1$
$$A=\int u^2(u^2+1)du$$ $$A=\int(u^4+u^2)du$$ $$A=\frac{u^5}{5}+\frac{u^3}{3}+C$$ $$A=\frac{\tan^5x}{5}+\frac{\tan^3x}{3}+C$$
