1. Calculate $\int 10 \sqrt[3]{x^2} \, dx$. Rewrite as $\int 10 x^{2/3} \, dx$. Integrate: $$10 \cdot \frac{x^{5/3}}{5/3} = 10 \cdot \frac{3}{5} x^{5/3} = 6 x^{5/3} + C.$$\n\n2. Calculate $\int \frac{3}{\sqrt{y}} \, dy = \int 3 y^{-1/2} \, dy$.\nIntegrate: $$3 \cdot \frac{y^{1/2}}{1/2} = 3 \cdot 2 y^{1/2} = 6 y^{1/2} + C.$$\n\n3. Calculate $\int 6 t^2 \sqrt[3]{t} \, dt = \int 6 t^{2 + \frac{1}{3}} dt = \int 6 t^{7/3} dt$.\nIntegrate: $$6 \cdot \frac{t^{10/3}}{10/3} = 6 \cdot \frac{3}{10} t^{10/3} = \frac{18}{10} t^{10/3} = \frac{9}{5} t^{10/3} + C.$$\n\n4. Calculate $\int (3 u^5 - 2 u^3) du = \int 3 u^5 du - \int 2 u^3 du$.\nIntegrate each: $$3 \cdot \frac{u^6}{6} - 2 \cdot \frac{u^4}{4} = \frac{1}{2} u^6 - \frac{1}{2} u^4 + C.$$\n\n5. Calculate $\int y^3 (2 y^2 - 3) dy = \int (2 y^5 - 3 y^3) dy$.\nIntegrate: $$2 \cdot \frac{y^6}{6} - 3 \cdot \frac{y^4}{4} = \frac{1}{3} y^6 - \frac{3}{4} y^4 + C.$$\n\n6. Calculate $\int \frac{x^2 + 4 x - 4}{\sqrt{x}} dx = \int (x^2 + 4 x - 4) x^{-1/2} dx = \int (x^{3/2} + 4 x^{1/2} - 4 x^{-1/2}) dx$.\nIntegrate terms: $$\frac{2}{5} x^{5/2} + \frac{8}{3} x^{3/2} - 8 x^{1/2} + C.$$\n\n7. Calculate $\int (3 \sin t - 2 \cos t) dt = 3 \int \sin t dt - 2 \int \cos t dt = -3 \cos t - 2 \sin t + C.$\n\n8. Calculate $\int (\sqrt{3} x + 4) dx = \sqrt{3} \int x dx + 4 \int dx = \frac{\sqrt{3}}{2} x^2 + 4 x + C.$\n\n9. Calculate $\int x^2 (5 - 2 x^3) dx = \int (5 x^2 - 2 x^5) dx$.\nIntegrate: $$\frac{5}{3} x^3 - \frac{2}{6} x^6 = \frac{5}{3} x^3 - \frac{1}{3} x^6 + C.$$\n\n10. Calculate $\int y^3 \sqrt{1 + y} dy = \int y^3 (1 + y)^{1/2} dy$. Set $u = 1 + y$, $y = u - 1$, $dy = du$.\nRewrite: $$\int (u - 1)^3 u^{1/2} du = \int (u^3 - 3 u^2 + 3 u - 1) u^{1/2} du = \int (u^{7/2} - 3 u^{5/2} + 3 u^{3/2} - u^{1/2}) du.$$\nIntegrate each term and substitute back: $$\frac{2}{9} (1 + y)^{9/2} - \frac{6}{7} (1 + y)^{7/2} + \frac{6}{5} (1 + y)^{5/2} - \frac{2}{3} (1 + y)^{3/2} + C.$$\n\n11. Calculate $\int \frac{\sin \theta}{\cos^2 \theta} d\theta$. Use substitution $u = \cos \theta$, $du = -\sin \theta d\theta$,\nIntegral becomes $$- \int u^{-2} du = - \left(- u^{-1} \right) + C = \frac{1}{\cos \theta} + C = \sec \theta + C.$$\n\n12. Calculate $\int \sin t \sqrt{1 - \cos t} dt$. Set $u = 1 - \cos t$, $du = \sin t dt$, integral becomes $$\int \sqrt{u} du = \frac{2}{3} u^{3/2} + C = \frac{2}{3} (1 - \cos t)^{3/2} + C.$$\n\n13. Calculate $\int \frac{s}{\sqrt{3 s^2 + 1}} ds$. Set $u = 3 s^2 + 1$, $du = 6 s ds$, $s ds = \frac{du}{6}$, integral becomes $$\int \frac{s}{\sqrt{u}} ds = \frac{1}{6} \int u^{-1/2} du = \frac{1}{6} \cdot 2 u^{1/2} + C = \frac{\sqrt{3 s^2 + 1}}{3} + C.$$\n\n14. Calculate $\int \cos (4 \theta) d\theta = \frac{\sin (4 \theta)}{4} + C.$\n\n15. Calculate definite integral $\int_{-1}^2 (5 - 2 x) dx$. Antiderivative: $5 x - x^2$. Evaluate at bounds:\nAt 2: $5 \cdot 2 - 2^2 = 10 - 4 = 6$.\nAt -1: $5 \cdot (-1) - (-1)^2 = -5 - 1 = -6$.\nResult: $6 - (-6) = 12$.\n\n16. Calculate $\int_0^\pi (2 \sin t + 3 \cos t + 1) dt$.\nAntiderivative: $-2 \cos t + 3 \sin t + t$. Evaluate:\n$[-2 \cos \pi + 3 \sin \pi + \pi] - [-2 \cos 0 + 3 \sin 0 + 0] = [-2(-1) + 0 + \pi] - [-2(1) + 0 + 0] = (2 + \pi) - (-2) = \pi + 4.$\n\n17. Calculate definite integral $\int_2^7 s^2 \sqrt{1 + s} ds$. Set $u = 1 + s$, then $s = u - 1$, change limits from $s=2$ to $u=3$, $s=7$ to $u=8$, integral becomes $$\int_3^8 (u - 1)^2 u^{1/2} du = \int_3^8 (u^2 - 2 u + 1) u^{1/2} du = \int_3^8 (u^{5/2} - 2 u^{3/2} + u^{1/2}) du.$$\nIntegrate: $$\frac{2}{7} u^{7/2} - \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} \Big|_3^8.$$\nEvaluate numerically if needed.\n\n18. Calculate definite integral $\int_0^{\frac{\pi^2}{4}} \frac{\sin \sqrt{x}}{\sqrt{x}} dx$. Use substitution $t = \sqrt{x}$, $x = t^2$, $dx = 2 t dt$, integral becomes $$\int_0^{\frac{\pi}{2}} \frac{\sin t}{t} 2 t dt = 2 \int_0^{\frac{\pi}{2}} \sin t dt = 2 [-\cos t]_0^{\pi/2} = 2 (1 - 0) = 2.$$