Question

Інтеграл від 0 до п/6 sin 2x dx
Інтеграл від 0 до п/6 2 сos x dx

Answer 1

$\int\limits_{0}^{\frac{\pi}{6}} \sin 2x\,dx = -\dfrac{\cos 2x}{2}\ \ \Bigg|_0^{\frac{\pi}{6}}\ = -\dfrac{\cos\left(2\cdot \dfrac{\pi}{6}\right)}{2} - \left(-\dfrac{\cos(2\cdot 0)}{2}\right) =\\\\\\= -\dfrac{\cos\dfrac{\pi}{3}}{2} + \dfrac{\cos 0}{2} = -\dfrac{\dfrac{1}{2}}{2} + \dfrac{1}{2} = -\dfrac{1}{4} + \dfrac{1}{2} = \dfrac{1}{4} = \bf{0,25}$

$\int\limits_0^{\frac{\pi}{6}}2\cos x\,dx = 2\sin x\ \ \Bigg|_0^{\frac{\pi}{6}}\ = 2\sin \dfrac{\pi}{6} - 2\sin 0 = 2\cdot \dfrac{1}{2} - 2\cdot 0 = 1 - 0 = \bf{1}$