Question

помогите разобраться...........

Answer 1

$1)\; \; \int\limits^{\frac{\pi}{2}}_0\, sin2x\cdot cos3x\, dx= \int\limits^{\frac{\pi}{2}}_0\, \frac{1}{2}\cdot (sin5x+sin(-x))dx=\\\\=\frac{1}{2}\cdot \int\limits^{\frac{\pi}{2}}_0(sin5x-sinx)dx=\frac{1}{2}\cdot (-\frac{1}{5}cos5x+cosx)\Big |_0^{\frac{\pi }{2}}=\\\\=\frac{1}{2}\cdot (-\frac{1}{5}cos\frac{5\pi }{2}+cos\frac{\pi}{2}+\frac{1}{5}cos0-cos0)=\\\\=\frac{1}{2}\cdot (-\frac{1}{5}\cdot 0+0+\frac{1}{5}\cdot 1-1)=\frac{1}{2}\cdot \frac{-4}{5}=-\frac{2}{5}$

$2)\; \; \int\limits^{\pi }_{\frac{\pi}{4}}\, cos^2\frac{x}{2}\, dx= \int\limits^{\pi }_{\frac{\pi}{4}}\frac{1+cosx}{2}\, dx =\frac{1}{2}\cdot \int\limits^{\pi }_{\frac{\pi }{4}} (1+cosx)\, dx=\\\\=\frac{1}{2}\cdot (x+sinx)\Big |_{\frac{\pi}{4}}^{\pi }=\frac{1}{2}\cdot \Big (\pi +sin\pi -\frac{\pi }{4}-sin\frac{\pi }{4}\Big )=\\\\=\frac{1}{2}\cdot \Big (\frac{3\pi }{4}- \frac{\sqrt2}{2} \Big )=\frac{3\pi }{8}- \frac{\sqrt2}{4}= \frac{3\pi -2\sqrt2}{8}$