## University Calculus: Early Transcendentals (3rd Edition)

$$\int\tan^5xdx=\frac{\sec^4x}{4}-\tan^2x+\ln|\sec x|+C$$
$$A=\int\tan^5xdx=\int\tan x(\tan^2x)^2dx$$ $$A=\int\tan x(\sec^2x-1)^2dx$$ $$A=\int\tan x(\sec^4x-2\sec^2x+1)dx$$ $$A=\int\tan x\sec^4xdx-\int2\tan x\sec^2xdx+\int\tan xdx$$ $$A=X-Y+Z$$ 1) For $X$: $$X=\int\tan x\sec^4xdx=\int\sec^3x(\tan x\sec x)dx$$ $$X=\int\sec^3xd(\sec x)$$ $$X=\int u^3du=\frac{u^4}{4}+C$$ (set $u=\sec x$) $$X=\frac{\sec^4x}{4}+C$$ 2) For $Y$: $$Y=\int2\tan x\sec^2xdx=\int2\tan xd(\tan x)$$ $$Y=\int2udu$$ (set $u=\tan x$) $$Y=\frac{2u^2}{2}+C=u^2+C$$ $$Y=\tan^2x+C$$ 3) For $Z$: $$Z=\int\tan xdx$$ $$Z=\ln|\sec x|+C$$ Therefore, $$A=\frac{\sec^4x}{4}-\tan^2x+\ln|\sec x|+C$$