## Calculus: Early Transcendentals 8th Edition

$$\int\tan^2\theta\sec^4\theta d\theta=\frac{\tan^3\theta}{3}+\frac{\tan^5\theta}{5}+C$$
$$A=\int\tan^2\theta\sec^4\theta d\theta$$ In encounter with an integral with $\tan x$ and $\sec x$, 3 basic formulas need to be remembered: $$(\tan x)'=\sec^2 x$$ $$(\sec x)'=\sec x\tan x$$ $$\sec^2 x=1+\tan^2 x$$ We would use them for transformations as follows $$A=\int(\tan^2\theta\sec^2\theta)sec^2\theta d\theta$$ $$A=\int[\tan^2\theta(1+\tan^2\theta)]sec^2\theta d\theta$$ *To do the Substitution here, we would let $u=\tan\theta$ Let $u=\tan\theta$. Then $du=\sec^2\theta d\theta$. Which means, $$A=\int u^2(1+u^2)du$$ $$A=\int(u^2+u^4)du$$ $$A=\frac{u^3}{3}+\frac{u^5}{5}+C$$ $$A=\frac{\tan^3\theta}{3}+\frac{\tan^5\theta}{5}+C$$