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$\int\limits^4_{-2} {3\sqrt{3-\frac{x}{3}}-\frac{81}{(2x-5)^3}} \, dx=\\ \int\limits^4_{-2} {3\sqrt{3-\frac{x}{3}}} \, dx-\int\limits^4_{-2} {\frac{81}{(2x-5)^3}} \, dx=\\ 9\int\limits^4_{-2} {\sqrt{3-\frac{x}{3}}} \, d(3-\frac{x}{3})-\frac{81}{2}\int\limits^4_{-2} {(2x-5)^{-3}} \, d(2x-5)=\\ 9*\frac{(3-\frac{x}{3})^{\frac{3}{2}}}{\frac{3}{2}}|^4_{-2}-\frac{81}{2}\frac{{(2x-5)^{-2}}}{-2}|^4_{-2}=\\ 18(3-\frac{x}{3})^{\frac{3}{2}}|^4_{-2}+\frac{81(2x-5)^{-2}}{4}|^4_{-2}=\\$

$18*(3-\frac{4}{3})^{\frac{3}{2}}-18*(3-\frac{-2}{3})^{\frac{3}{2}}+\frac{81(2*4-5)^{-2}}{4}-\frac{81(2*(-2)-5)^{-2}}{4}=\\ 18*\frac{5}{3}*\sqrt {\frac{5}{3}}-18*\frac{11}{3}\sqrt{\frac{11}{3}}+\frac{9}{4}-\frac{1}{4}=\\ 30*\sqrt {\frac{5}{3}}-66*\sqrt{\frac{11}{3}}+2$

$\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{8}} {(sin^2(2x)-\frac{1}{2})} \, dx=\\ \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{8}} {\frac{1}{2}-cos(2* 2x)-\frac{1}{2})} \, dx=\\ \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{8}} {cos(4x)} \, dx=\\ \frac{sin (4x)}{4} |^{\frac{\pi}{2}}_{\frac{\pi}{8}}=\\ \frac{sin (4*\frac{\pi}{2})}{4}-\frac{sin (4*\frac{\pi}{8})}{4}=\\ \frac{sin (2\pi)}{4}-\frac{sin (\frac{\pi}{2})}{4}=\\ 0-\frac{1}{4}=-0.25$