Question

Вычислите интеграл [п/4; 0] ∫ 2sin4x dx

Answer 1

$\displaystyle \int\limits^\big{\frac{\pi}{4}}_02\sin4x\,dx=2\cdot\bigg(-\dfrac{\cos4x}{4}\bigg)\bigg|^\big{\frac{\pi}{4}}_0=-\dfrac{\cos4x}{2}\bigg|^\big{\frac{\pi}{4}}_0=-\dfrac{\cos(4\cdot\frac{\pi}{4})}{2}-\\ \\ \\ -\bigg(-\dfrac{\cos(4\cdot0)}{2}\bigg)=-\dfrac{\overbrace{\cos\pi}^{-1}}{2}+\dfrac{\overbrace{\cos0}^{1}}{2}=\dfrac{1}{2}+\dfrac{1}{2}=1$