Question

Интеграл. Вычислите 1) ∫(вверху 2П внизу -П) sin x/3 dx 2) ∫(вверху П/2 внизу 0) 4dx/cos^2 x/3 3) ∫(вверху П/3 внизу П/6) dx/sin^2 4x

Answer 1

$1)\; \; \int\limits^{2\pi }_{-\pi }\, sin\frac{x}{3}\, dx =-3cos \frac{x}{3}\Big |_{-\pi } ^{2\pi }=-3(cos\frac{2\pi}{3}-cos\frac{\pi}{3})=-3(-0,5-0,5)=3\\\\2)\; \; \int\limits^{\frac{\pi}{2}}_0\, \frac{4\, dx}{cos^2\frac{x}{3}} \, dx =4\cdot 3\cdot tg\frac{x}{3}\Big |_0^{\frac{\pi}{2}}=12(tg\frac{\pi}{6}-tg0)=12\, \frac{\sqrt3}{3}=4\sqrt3$

$3)\; \; \int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\, \frac{dx}{sin^24x} =- \frac{1}{4}\cdot ctg4x\Big |_{\frac{\pi}{6}} ^\frac{\pi}{3}=- \frac{1}{4}\, (ctg\frac{4\pi}{3}-ctg\frac{2\pi}{3})=\\\\=-\frac{1}{4}\, (ctg \frac{\pi }{3} -ctg\frac{2\pi }{3} )=-\frac{1}{4}\, ( \frac{\sqrt3}{3}+ \frac{\sqrt3}{3})=- \frac{\sqrt3}{6}$