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Решите задачу:

$\Large \int_{-2}^{1}{\mathrm dx \over (11+5x)^3}={1\over5}\int_{-2}^{1}{\mathrm d(11+5x) \over (11+5x)^3}={1\over5}{(11+5x)^{-2}\over -2}|_{-2}^{1}=-{1\over10}{1\over(11+5x)^2}|_{-2}^{1}=-{1\over10}\left ( {1\over16^2} -{1\over 1^2}\right )=-{1\over10}\cdot{-255\over256}={255\over2560}={51\over512}\\\\ \int_{0}^{\pi\over2}\cos^5{x}\sin{(2x)}\mathrm dx=2\int_{0}^{\pi\over2}\cos^6{x}\sin{x}\mathrm dx=-2\int_{0}^{\pi\over2}\cos^6{x}\mathrm d{(\cos{x})}={-2\over7}\cos^7{x}|_{0}^{\pi\over2}={-2\over7}\left ( cos^7{\left (\pi\over2 \right )}-cos^7{(0)} \right )={2\over7}$