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

$$\int8\cos^42\pi xdx=3x+\frac{\sin4\pi x}{\pi}+\frac{\sin8\pi x}{8\pi}+C$$
$$A=\int8\cos^42\pi xdx$$ Set $a=2\pi x$, which means $$da=2\pi dx$$ $$dx=\frac{1}{2\pi}da$$ Therefore, $$A=\int\frac{8}{2\pi}\cos^4ada=\int\frac{4}{\pi}\cos^4ada$$ $$A=\frac{4}{\pi}\int(\cos^2a)^2da$$ This is Case 3, so we need to rewrite $$\cos^2a=\frac{1+\cos2a}{2}$$ We have, $$A=\frac{4}{\pi}\int\Big(\frac{1+\cos2a}{2}\Big)^2da$$ $$A=\frac{1}{\pi}\int(1+\cos2a)^2da$$ $$A=\frac{1}{\pi}\int(1+2\cos2a+\cos^22a)da$$ $$A=\frac{1}{\pi}\Big(a+\sin2a\Big)+\frac{1}{\pi}\int\cos^22ada$$ The integral is again an example of Case 3, so we would rewrite $$\cos^22a=\frac{1+\cos4a}{2}$$ $$A=\frac{1}{\pi}\Big(a+\sin2a\Big)+\frac{1}{\pi}\int\frac{1+\cos4a}{2}da$$ $$A=\frac{1}{\pi}\Big(a+\sin2a\Big)+\frac{1}{\pi}\Big(\frac{a}{2}+\frac{\sin4a}{8}\Big)+C$$ $$A=\frac{1}{\pi}\Big(\frac{3a}{2}+\sin2a+\frac{\sin4a}{8}\Big)+C$$ $$A=\frac{1}{\pi}\Big(\frac{6\pi x}{2}+\sin4\pi x+\frac{\sin8\pi x}{8}\Big)+C$$ $$A=3x+\frac{\sin4\pi x}{\pi}+\frac{\sin8\pi x}{8\pi}+C$$