Question

$\int\limits^ \frac{ \pi }{2} _ \frac{ \pi }{6} {cos4x} \, dx$

Answer 1

$\int _{\frac{\pi}{6}}^{\frac{\pi}{2}}\, cos4x\, dx=[\, t=4x,\; dt=4\, dx,\; dx=\frac{dt}{4},\\\\ \int cos4x\, dx=\frac{1}{4}\int cost\, dt=\frac{1}{4}sint+C=\frac{1}{4}sin4x+C\, ]=\\\\=\frac{1}{4}sin4x\, |_{\frac{\pi}{6}}^{\frac{\pi}{2}}=\frac{1}{4}(sin2\pi -sin\frac{2\pi}{3})=\frac{1}{4}(0-\frac{\sqrt3}{2})=-\frac{\sqrt3}{8}$