Answer
$$\int 8\cos^42\pi tdt=\frac{\cos^32\pi t\sin2\pi t}{\pi}+\frac{3\sin4\pi t}{4\pi}+3t+C$$
Work Step by Step
$$A=\int 8\cos^42\pi tdt$$
Use Reduction Formula 68, which states that
$$\int \cos^nax=\frac{\cos^{n-1}ax\sin ax}{na}+\frac{n-1}{n}\int\cos^{n-2}axdx$$
for $a=2\pi$ and $n=4$
$$A=8\Big[\frac{\cos^32\pi t\sin2\pi t}{8\pi}+\frac{3}{4}\int\cos^22\pi tdt\Big]$$
Use Reduction Formula 68 one more time, this time for $a=2\pi$ and $n=2$
$$A=8\Big[\frac{\cos^32\pi t\sin2\pi t}{8\pi}+\frac{3}{4}\Big(\frac{\cos2\pi t\sin2\pi t}{4\pi}+\frac{1}{2}\int\cos^02\pi tdt\Big)\Big]$$ $$A=8\Big[\frac{\cos^32\pi t\sin2\pi t}{8\pi}+\frac{3}{4}\Big(\frac{\frac{1}{2}\sin4\pi t}{4\pi}+\frac{1}{2}\int 1dt\Big)\Big]$$ $$A=8\Big[\frac{\cos^32\pi t\sin2\pi t}{8\pi}+\frac{3}{4}\Big(\frac{\sin4\pi t}{8\pi}+\frac{t}{2}\Big)\Big]+C$$ $$A=\frac{\cos^32\pi t\sin2\pi t}{\pi}+6\Big(\frac{\sin4\pi t}{8\pi}+\frac{t}{2}\Big)+C$$ $$A=\frac{\cos^32\pi t\sin2\pi t}{\pi}+\frac{3\sin4\pi t}{4\pi}+3t+C$$
