помогите..............................

$\int\limits^2_{-1} {((1-x)^4)-\frac{1}{\sqrt{5x+6}}} \, dx=\\ \int\limits^2_{-1} {(1-x)^4} \, dx-\int\limits^2_{-1} {\frac{1}{\sqrt{5x+6}}} \, dx=\\ -\int\limits^2_{-1} {(1-x)^4} \, d(1-x)-\frac{1}{5}\int\limits^2_{-1} {\frac{1}{\sqrt{5x+6}}} \, d(5x+6)=\\ -\frac{(1-x)^5}{5}|^2_{-1}-\frac{1}{5}*\frac{\sqrt{5x+6}}{0.5}|^2_{-1}=\\ -\frac{(1-2)^5}{5}+\frac{(1-(-1))^5}{5}-\frac{1}{5}*\frac{\sqrt{5*2+6}}{0.5}+\frac{1}{5}*\frac{\sqrt{5*(-1)+6}}{0.5}=\\ \frac{1}{5}+\frac{32}{5}-\frac{8}{5}+\frac{2}{5}=6.6-1.2=5.4$

$\int\limits^{\frac{\pi}{4}}_0 {sin(2x)cos(2x)} \, dx=\\ \frac{1}{2}\int\limits^{\frac{\pi}{4}}_0 {2sin(2x)cos(2x)} \, dx=\\ \frac{1}{2}\int\limits^{\frac{\pi}{4}}_0 {sin(2*2x)} \, dx=\\ \frac{1}{2}\int\limits^{\frac{\pi}{4}}_0 {sin(4x)} \, dx=\\ \frac{1}{8}\int\limits^{\frac{\pi}{4}}_0 {sin(4x)} \, d(4x)=\\ \frac{1}{8}*(-cos(4x))|^{\frac{\pi}{4}}_0=\\ \frac{1}{8}*(-cos(4*\frac{\pi}{4}))-\frac{1}{8}*(-cos(4*0))=\\ 0.125+0.125=0.25$