V = int (0 1) (arcsin x)^2 dx
int (arcsin x)^2 dx = | arcsin x = t; x = sin t; dx = d(sin t) | =
= int t^2 d(sin t) = t^2 * sin t -int sin t d(t^2) = t^2 * sin t - 2 * int t * sin t dt =
= t^2 * sin t + 2 * int t d(cos t) = t^2 * sin t + 2 * (t * cos t - int cos t dt) =
= t^2 * sin t + 2 * t * cos t - 2 * sin t +C = | x = sin t; arcsin x = t | =
= (arcsin x)^2 * x + 2 * arcsin x * cos (arcsin x) - 2 * x +C =
= (arcsin x)^2 * x + 2 * arcsin x * (1 - x^2)^(1/2) - 2 * x + C
Тогда
V = int (0 1) (arcsin x)^2 dx =
= ((arcsin x)^2 * x + 2 * arcsin x * (1 - x^2)^(1/2) - 2 * x)_{0}^{1} =
= ((arcsin 1)^2 * 1 + 2 * arcsin 1 * (1 - 1^2)^(1/2) - 2 * 1) -
- ((arcsin 0)^2 * 0 + 2 * arcsin 0 * (1 - 0^2)^(1/2) - 2 * 0) =
= (pi/2)^2 - 2 = pi^2/4 - 2
Ответ: V = pi^2/4 - 2.