Question

∫√(1-х)arcsin√(x) dx

Answer 1

$\int \sqrt{1-x}\cdot arcsin\sqrt{x}\, dx=\Big [\, u=arcsin\sqrt{x},\\\\du=\frac{\frac{1}{2\sqrt{x}}dx}{\sqrt{1-(\sqrt{x})^2}}=\frac{dx}{2\sqrt{x}\cdot \sqrt{1-x}},\; dv=\sqrt{1-x}dx,\; v=\int \sqrt{1-x}dx=\\\\=\int (1-x)^{1/2}dx=-\frac{(1-x)^{3/2}}{3/2}=-\frac{2(1-x)^{3/2}}{3}\, \Big ]=uv-\int v\, du=\\\\=-\frac{2(1-x)^{3/2}}{3}\cdot arcsin\sqrt{x}+\frac{2}{3}\int \frac{(1-x)^{3/2}}{2\sqrt{x}\cdot \sqrt{1-x}}=\\\\=-\frac{2\sqrt{(1-x)^3}}{3}\cdot arcsin\sqrt{x}+\frac{1}{3}\int \frac{1-x}{\sqrt{x}}dx=$

$=-\frac{2\sqrt{(1-x)^2}}{3}\cdot arcsin\sqrt{x}+\frac{1}{3}\cdot (\int\frac{dx}{\sqrt{x}}-\int \sqrt{x}\, dx)=\\\\=-\frac{2\sqrt{(1-x)^3}}{3}\cdot arcsin\sqrt{x}+\frac{1}{3} \cdot (2\sqrt{x}-\frac{2\sqrt{x^3}}{3})+C$