Question

Como resolvo essa integral?
$\int\limits^a_b {3 cos ^{2} 2x } \, dx$
sendo a=π/4
b=π/8

Answer 1

Sabemos que:

$\cos^2(a)=\dfrac{\cos(2a)+1}{2}$

Entao:

$\int\limits^\frac{\pi}{4}_\frac{\pi}{8} {3\cos^2(2x)} \, dx =\\\\\\3 \int\limits^\frac{\pi}{4}_\frac{\pi}{8} {\dfrac{\cos(4x)+1}{2}} \, dx=\\\\\\\dfrac32 \int\limits^\frac{\pi}{4}_\frac{\pi}{8} {\cos(4x)+1} \, dx=\\\\\\\dfrac32\left[\dfrac{\sin(4x)}{4}+x\right]^\frac{\pi}{4}_\frac{\pi}{8}=\\\\\\\dfrac32\left[\dfrac{\sin(\pi)}{4}+\dfrac{\pi}{4}-\left(\dfrac{\sin(\frac{\pi}{2})}{4}+\dfrac{\pi}{8}\right)\right]=\\\\\\\dfrac32\left[0+\dfrac\pi4-\dfrac14-\dfrac\pi8\right]=\dfrac32\left[\dfrac{\pi-2}{8}\right]=$

$\boxed{\dfrac{3\pi-6}{16}}$