Answer
$$\int\sec^6xdx=\frac{\tan^5x}{5}+\frac{2\tan^3x}{3}+\tan x+C$$
Work Step by Step
$$A=\int\sec^6xdx$$ $$A=\int(\sec^2x)^2(\sec^2xdx)$$ $$A=\int(\tan^2x+1)^2d(\tan x)$$
We set $u=\tan x$.
$$A=\int(u^2+1)^2du$$ $$A=\int(u^4+2u^2+1)du$$ $$A=\frac{u^5}{5}+\frac{2u^3}{3}+u+C$$ $$A=\frac{\tan^5x}{5}+\frac{2\tan^3x}{3}+\tan x+C$$
