Question

Вычислить определенный интеграл

Answer 1

Ответ:

$\displaystyle \int\limits^{\pi /2}_ {-\pi /2}\, \sqrt{cosx-cos^3x}\, dx=\int\limits^{\pi /2}_ {-\pi /2}\, \sqrt{cosx\, (1-cos^2x)}\, dx=\int\limits^{\pi /2}_ {-\pi /2}\, \sqrt{cosx\cdot sin^2x}\, dx=\\\\\\=\int\limits^{\pi /2}_ {-\pi /2}\, \sqrt{cosx}\cdot \underbrace {sinx\, dx}_{-d\, (cosx)}=-\int\limits^{\pi /2}_ {-\pi /2}\, \sqrt{cosx}\cdot d\, (cosx)=-\frac{(cosx)^{3/2}}{3/2}\Big|_{-\pi /2}^{\pi /2}=\\\\\\=-\frac{2\sqrt{cos^3x}}{3}\, \Big|_{-\pi /2}^{\pi /2}=-\frac{2}{3}\cdot (\sqrt{0^3}-\sqrt{0^3})=0$