Question

С 472(4) до конца, решите кто может пожалуйста

Answer 1

$4)\; \int\limits^0_{-\sqrt[3]3}\, x^2\cdot e^{\frac{x^3}{3}+1}\, dx=[\; t=\frac{x^3}{3}+1\; ,\; dt=x^2\, dx\; ,\; \int x^2\cdot e^{\frac{x^3}{3}+1}\, dx=\\\\=\int e^{t}\, dt=e^{t}+C\; ]=e^{\frac{x^3}{3}+1}\, \Big |^0_{-\sqrt[3]3}=e-e^{0}=e-1\; ;\\\\\\6)\; \int\limits^0_{-1/\sqrt2}\, \frac{x\, dx}{5\, \sqrt{1-x^4}} dx=\int\limits^0_{-1/\sqrt2}\, \frac{2x\, dx}{10\sqrt{1-(x^2)^2}}=\int\limits^0_{-1\sqrt2}\, \frac{d(x^2)}{10\sqrt{1-(x^2)^2}}=\\\\=\frac{1}{10}\cdot arcsin(x^2)\Big |_{-1/\sqrt2}^0=\frac{1}{10}\cdot (arcsin0-arcsin\frac{1}{2})=\frac{1}{10}\cdot \frac{\pi }{6}=\frac{\pi}{60}$

$5)\; \; \int\limits^{\sqrt{\pi /2}}_{\sqrt{\pi }/2}\, \frac{x\cdot sinx^2}{2}\, dx= \int\limits^{\sqrt{\pi /2}}_{\sqrt{\pi }/2}\, \frac{2x\cdot sinx^2}{4}\, dx=\frac{1}{4} \int\limits^{\sqrt{\pi /2}}_{\sqrt{\pi }/2}\, sinx^2\cdot d(x^2)=\\\\=\frac{1}{4}\cdot (-cosx^2)\Big |_{\sqrt{\pi}/2}^{\sqrt{\pi /2}}=-\frac{1}{4}\cdot (cos\frac{\pi}{2}-cos\frac{\pi}{4})=-\frac{1}{4}\cdot (-\frac{\sqrt2}{2})=\frac{\sqrt2}{8}$