## Calculus: Early Transcendentals 8th Edition

$$\int\tan x\sec^3 xdx=\frac{\sec^3 x}{3}+C$$
$$A=\int\tan x\sec^3 xdx$$ $$A=\int(\tan x\sec x)\sec^2 xdx$$ In encounter with an integral with $\tan x$ and $\sec x$, 3 basic formulas need to be remembered: $$(\tan x)'=\sec^2 x$$ $$(\sec x)'=\sec x\tan x$$ $$\sec^2 x=1+\tan^2 x$$ *Here we see that if we put $u=\sec x$, then $du=\sec x\tan xdx$. That means we are left only with $u^2$ after the Substitution Rule. So that seems the wisest way to do the Substitution. Let $u=\sec x$. Then $du=\sec x\tan xdx$. Therefore, $$A=\int u^2du$$ $$A=\frac{u^3}{3}+C$$ $$A=\frac{\sec^3 x}{3}+C$$ (There is generally not only one way to deal with these kinds of trigonometrical integrals. Some require more transformations and even a combination of several methods and therefore, are harder than the others. So it is wiser to find the optimal way to deal with them.)