Решите пожалуйста. Все три

1-й пример:
$5x+1 \equiv t, \ \int {5\sqrt{t}} \, \cdot dt = \\ =5\int {\sqrt{t}} \, \cdot dt = 5 \cdot {\frac{2\sqrt{t^3}}{3}} = {\frac{10\sqrt{t^3}}{3}} = {\frac{10\sqrt{(5x+1)^3}}{3}} = \\ \int\limits^3_0 {\sqrt{5x+1}} \, dx = {\frac{10\sqrt{(5x+1)^3}}{3}} |^{3}_{0}= \\ ={\frac{10\sqrt{(5 \cdot 3+1)^3}}{3}} - {\frac{10\sqrt{(5 \cdot 0 +1)^3}}{3}} = \frac {640}{3}-\frac{10}{3}=\\ =\frac{630}{3}=210$
2-й пример:
$\int {cos \ 2x} \, dx = \frac {sin \ 2x}{2} \\ \int\limits^{\frac{ \pi }{3}}_{-\frac{\pi}{6}} {cos \ 2x} \, dx = \frac {sin \ 2x}{2} |^{\frac{ \pi }{3}}_{-\frac{\pi}{6}} = \\ = \frac {sin \ \frac{ 2 \cdot \pi }{3}}{2} - \frac {sin \ \frac{ 2 \cdot \pi }{6}}{2} = \\ = \frac {-1}{2} - \frac {\frac {\sqrt{3}}{2}}{2} = -\frac{2+\sqrt{3}}{4}$
3-й пример:
$\int {x \cdot e^x} \, dx = e^x \cdot (x-1) +C \\ \int\limits^1_0 {x \cdot e^x} \, dx = e^x \cdot (x-1)|^{1}_{0} = e^1 \cdot (1-1) - e^0 \cdot (0-1) = \\ = e \cdot 0 - 1 \cdot (-1) = 0 + 1 = 1$