Вычислите определенные интегралы: 6. а) ∫(0 to π/4) (4 cos 2x)dx, б) ∫(1 to 5) √(9x-1)dx; 7. а) ∫(π/2 to π) (6 sin 2x)dx, б) ∫(1 to 2) (3-2x)⁴dx; 8. а) ∫(0 to π/3) 5/cos²x dx, б) ∫(-3 to 1) dx/(5-3x).

6. а) $\int_{0}^{\frac{\pi}{4}} (4 \cos 2x) dx = 4 \cdot \frac{1}{2} \sin 2x \Big|_0^{\frac{\pi}{4}} = 2 \sin(2 \cdot \frac{\pi}{4}) - 2 \sin 0 = 2 \sin \frac{\pi}{2} - 0 = 2 \cdot 1 = 2$ Ответ: 2 б) $\int_{1}^{5} \sqrt{9x - 1} dx = \int_{1}^{5} (9x - 1)^{\frac{1}{2}} dx = \frac{1}{9} \cdot \frac{(9x - 1)^{\frac{3}{2}}}{\frac{3}{2}} \Big|_1^5 = \frac{2}{27} (9x - 1)\sqrt{9x - 1} \Big|_1^5 = \frac{2}{27} (44\sqrt{44} - 8\sqrt{8}) = \frac{2}{27} (88\sqrt{11} - 16\sqrt{2})$ Ответ: $\frac{16}{27} (11\sqrt{11} - 2\sqrt{2})$ 7. а) $\int_{\frac{\pi}{2}}^{\pi} (6 \sin 2x) dx = 6 \cdot (-\frac{1}{2} \cos 2x) \Big|_{\frac{\pi}{2}}^{\pi} = -3 \cos 2x \Big|_{\frac{\pi}{2}}^{\pi} = -3(\cos 2\pi - \cos \pi) = -3(1 - (-1)) = -3 \cdot 2 = -6$ Ответ: -6 б) $\int_{1}^{2} (3 - 2x)^4 dx = -\frac{1}{2} \cdot \frac{(3 - 2x)^5}{5} \Big|_1^2 = -\frac{1}{10} (3 - 2 \cdot 2)^5 - (-\frac{1}{10} (3 - 2 \cdot 1)^5) = -\frac{1}{10}(-1)^5 + \frac{1}{10}(1)^5 = \frac{1}{10} + \frac{1}{10} = 0,2$ Ответ: 0,2 8. а) $\int_{0}^{\frac{\pi}{3}} \frac{5}{\cos^2 x} dx = 5 \tan x \Big|_0^{\frac{\pi}{3}} = 5 \tan \frac{\pi}{3} - 5 \tan 0 = 5\sqrt{3} - 0 = 5\sqrt{3}$ Ответ: $5\sqrt{3}$ б) $\int_{-3}^{1} \frac{dx}{5 - 3x} = -\frac{1}{3} \ln |5 - 3x| \Big|_{-3}^1 = -\frac{1}{3} (\ln |5 - 3| - \ln |5 + 9|) = -\frac{1}{3} (\ln 2 - \ln 14) = -\frac{1}{3} \ln \frac{2}{14} = -\frac{1}{3} \ln \frac{1}{7} = \frac{1}{3} \ln 7$ Ответ: $\frac{1}{3} \ln 7$