Question

Решите интегрлаьное исчесление

Answer 1

$\int \dfrac{sinx\, dx}{\sqrt[3]{cosx+1}}=\Big[\; t=cosx+1\; ,\; dt=-sinx\, dx\; \Big]=-\int \dfrac{dt}{\sqrt[3]{t}}=\\\\\\=-\int t^{-1/3}\, dt=-\dfrac{t^{2/3}}{2/3}+C=-\dfrac{3}{2}\, \sqrt[3]{(cosx+1)^2}+C\\\\\\\\\star \; \; \Big(-\dfrac{3}{2}\, \sqrt[3]{(cosx+1)^2}+C\Big)'=-\dfrac{3}{2}\cdot \dfrac{2}{3}\cdot (cosx+1)^{-1/3}\cdot (-sinx) =\dfrac{sinx}{\sqrt[3]{cosx+1}}\\\\\\\\\boxed {\; \int x^{k}\, dx=\dfrac{x^{k+1}}{k+1}+C}$