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

$$\int\sec^3x\tan^3xdx=\frac{\sec^5x}{5}-\frac{\sec^3x}{3}+C$$
$$A=\int\sec^3x\tan^3xdx$$ $$A=\int(\sec^2x\tan^2x)(\sec x\tan x)dx$$ Take $u=\sec x$, which means $$du=\sec x\tan xdx$$ Also, $\sec^2x=u^2$ and since $\sec^2x=\tan^2x+1$, we would have $\tan^2x=u^2-1$ $$A=\int u^2(u^2-1)du$$ $$A=\int(u^4-u^2)du$$ $$A=\frac{u^5}{5}-\frac{u^3}{3}+C$$ $$A=\frac{\sec^5x}{5}-\frac{\sec^3x}{3}+C$$