Предмет: Алгебра, автор: olesyaalexozggj4

Помогите 38 срочнооо

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Ответы

Автор ответа: NNNLLL54

$1)\; \; \int\limits^{\pi /4}_0\, (1-2cos^2x)\, dx=\int\limits^{\pi /4}\, (-cos2x)\, dx=-\frac{1}{2}\, sin2x\Big |_0^{\pi /4}=\\\\=-\frac{1}{2}\cdot (sin\frac{\pi }{2}-sin0)=-\frac{1}{2}\cdot (1-0)=-\frac{1}{2}\\\\\\2)\; \; \int\limits^{\pi /3}_0\, (2sin2x-1)\, dx=(2\cdot \frac{1}{2}\cdot (-cos2x)-x)\Big |_0^{\pi /3}=\\\\=-cos\frac{2\pi }{3}-\frac{\pi}{3}-cos0+0=\frac{1}{2}-\frac{\pi}{3}-1=-\frac{1}{2}-\frac{\pi}{3}$

$3)\; \; \int\limits^{\pi /4}_{\pi /6}\, (sinx+tgx\cdot ctgx)\, dx= \int\limits^{\pi /4}_{\pi /6}\, (sinx+1)\, dx=(-cosx+x)\Big |_{\pi /6}^{\pi /4}=\\\\=-cos\frac{\pi}{4}+\frac{\pi}{4}+cos\frac{\pi}{6}-\frac{\pi}{6}=-\frac{\sqrt2}{2}+\frac{\pi}{12}+\frac{\sqrt3}{2}=\frac{\sqrt3-\sqrt2}{2}+\frac{\pi}{12}$

$4)\; \; \int\limits^{\pi /2}_{\pi /3}\, (tg\frac{x}{5}\cdot ctg\frac{x}{5}-cos\frac{x}{5})\, dx=\int\limits^{\pi /2}_{\pi /3}\, (1-cos\frac{x}{5})\, dx=(x-5\, sin\frac{x}{5})\Big |_{\pi /3}^{\pi /2} =\\\\=\frac{\pi}{2}-5sin\frac{\pi}{10}-\frac{\pi}{3}+5sin\frac{\pi }{15}=\frac{\pi}{6}-5\, (sin\frac{\pi}{10}-sin\frac{\pi}{15})$