## Calculus: Early Transcendentals 8th Edition

$$\int\tan^5xdx=\frac{\sec^4 x}{4}-\sec^2 x+\ln|\sec x|+C$$
$$A=\int\tan^5xdx$$ $$A=\int\tan^4 x\tan xdx$$ We already know that $\tan^2 x=\sec^2x -1$ That means $\tan^4 x=(\sec^2 x-1)^2$ Therefore, $$A=\int(\sec^2 x-1)^2\tan xdx$$ $$A=\int(\sec^4 x-2\sec^2 x+1)\tan xdx$$ $$A=\int\sec^4 x\tan xdx-2\int\sec^2 x\tan xdx+\int\tan xdx$$ $$A=\int\sec^3 x(\sec x\tan x)dx-2\int\sec x(\sec x\tan x)dx+\ln|\sec x|+C$$ Now we do the Substitution. Let $u=\sec x$ Then $du=\sec x\tan xdx$. Therefore, $$A=\int u^3du-2\int udu+\ln|\sec x|+C$$ $$A=\frac{u^4}{4}-\frac{2u^2}{2}+\ln|\sec x|+C$$ $$A=\frac{u^4}{4}-u^2+\ln|\sec x|+C$$ $$A=\frac{\sec^4 x}{4}-\sec^2 x+\ln|\sec x|+C$$