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

$$\int\cos^34xdx=\frac{3\sin4x-(\sin4x)^3}{12}+C$$
$$A=\int\cos^34xdx$$ $$A=\frac{1}{4}\int\cos^34xd(4x)$$ We set $a=4x$. $$A=\frac{1}{4}\int\cos^3ada$$ Following the integral form $\int\sin^mx\cos^nxdx$, we would apply Case 2 here, where $m=0$ and $n=3$. $$A=\frac{1}{4}\int\cos^2a\cos ada$$ $$A=\frac{1}{4}\int(1-\sin^2a)d(\sin a)$$ We set $u=\sin a$. $$A=\frac{1}{4}\int(1-u^2)du$$ $$A=\frac{1}{4}\Big(u-\frac{u^3}{3}\Big)+C$$ $$A=\frac{1}{4}\Big(\frac{3u-u^3}{3}\Big)+C$$ $$A=\frac{3u-u^3}{12}+C$$ $$A=\frac{3\sin4x-(\sin4x)^3}{12}+C$$