Question

Интеграл 4*sqrt(x-x^2)dx

Answer 1

$\int 4\cdot \sqrt{x-x^2}\, dx=4\cdot \int \sqrt{-(x^2-x)}\, dx=4\int \sqrt{-(\, (x-\frac{1}{2})^2-\frac{1}{4}\, )}\, dx=\\\\=4\int \sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}\, dx=[\; t=x-\frac{1}{2}\; ,\; dt=dx\; ]=4\int \sqrt{\frac{1}{4}-t^2}\, dt=\\\\=4\int \sqrt{\frac{1-4t^2}{4}}\, dt=2\int \sqrt{1-4t^2}\, dt=[\; 2t=sinz\; ,\; 2\, dt=cosz\, dz\; ]=\\\\=\int \sqrt{1-sin^2z}\cdot cosz\, dz=\int \sqrt{cos^2z}\cdot cosz\, dz=\int cos^2z\, dz=\\\\=\int \frac{1+cos2z}{2}\, dz=\frac{1}{2}\int (1+cos2z)\, dz=\\\\=\frac{1}{2}\cdot (\, z+\frac{1}{2}\, sin2z\, )+C=\\\\=\frac{1}{2}\cdot (\, arcsin2t+\frac{1}{2}\, sin(2arcsin2t)\, )+C=\\\\=\frac{1}{2}\cdot (\, arcsin(2x-1)+\frac{1}{2}\cdot 2\, sin(arcsin2t)\cdot cos(arcsin2t)\, )+C=\\\\=\frac{1}{2}\cdot (\, arcsin(2x-1)+2t\cdot \sqrt{1-4t^2}\, )+C=\\\\=\frac{1}{2}\cdot (\, arcsin(2x-1)+(2x-1)\cdot \sqrt{4x-4x^2}\, )+C=\\\\=\frac{1}{2}\cdot (\, arcsin(2x-1)+2\cdot (2x-1)\cdot \sqrt{x-x^2}\, )+C\; ;$