Question

calculati integrala​

Answer 1

$\displaystyle I= \int_{0}^{2\pi}\arcsin\Big(\sin(2x)\Big)\, dx\\ \\\\\text{Folosesc formula lui King:}\\ \int_{a}^bf(x)\, dx = \int_{a}^b f(a+b-x)\, dx\\ \\ \\I = \int_{0}^{2\pi}\arcsin\Bigg(\sin\Big(2(0+2\pi - x)\Big)\Bigg)\, dx \\ \\ I =\int_{0}^{2\pi}\arcsin\Big(\sin(4\pi - 2x)\Big)\, dx \\ \\ I = \int_{0}^{2\pi}\arcsin\Big(\sin(-2x)\Big)\, dx\\ \\I = \int_{0}^{2\pi}-\arcsin\Big(\sin(2 x)\Big)\, dx \\ \\ I = -\int_{0}^{2\pi}\arcsin\Big(\sin(2 x)\Big)\, dx\\ \\ I = -I\\ \\ 2I = 0\\ \\ \Rightarrow \boxed{I = 0}$