Задача №1463

\[ \int\sin\sqrt[3]{x}dx =\left[\begin{aligned} & t=\sqrt[3]{x}; x=t^3.\\ & dx=3t^2dt. \end{aligned}\right] =3\int{t^2}\sin{t}dt =\left[\begin{aligned} & u=t^2;\;du=2tdt;\\ & dv=\sin{t}dt;\;v=-\cos{t}. \end{aligned}\right]=\\ =-3t^2\cos{t}+6\int{t}\cos{t}dt =\left[\begin{aligned} & u=t;\;du=dt;\\ & dv=\cos{t}dt;\;v=\sin{t}. \end{aligned}\right] =-3t^2\cos{t}+6t\sin{t}-6\int\sin{t}dt=\\ =-3t^2\cos{t}+6t\sin{t}+6\cos{t}+C =-3\sqrt[3]{x^2}\cos\sqrt[3]{x}+6\sqrt[3]{x}\sin\sqrt[3]{x}+6\cos\sqrt[3]{x}+C \]