Помогите, пожалуйста, вычислить интеграл
int (0 pi/2) sin x/(2 + sin x) dx

int (0 pi/2) sin x/(2 + sin x) dx = int (0 pi/2) (sin x + 2 - 2)/(sin x + 2) dx =
= int (0 pi/2) dx - 2 * int (0 pi/2) dx/(sin x + 2) =
= (x)_{0}^{pi/2} -
- 2 * int (0 pi/2) dx/(2 * sin (x/2) * cos (x/2) + 2 * sin^2 (x/2) + 2 * cos^2 (x/2)) dx =
= pi/2 - 2 * int (0 pi/2) dx/(cos^2 (x/2) * (2 * tg (x/2) + 2 * tg^2 (x/2) + 2)) dx =
= pi/2 - 4 * int (0 pi/2) d(tg (x/2))/(2 * tg (x/2) + 2 * tg^2 (x/2) + 2) dx =
= | tg (x/2) = t | = pi/2 - 4 * int (0 1) dt/(2 * t^2 + 2 * t + 2) =
= pi/2 - 2 * int (0 1) dt/(t^2 + t + 1) = pi/2 - 2 * int (0 1) dt/((t + 1/2)^2 + 3/4) =
= pi/2 - 2 * int (0 1) d(t + 1/2)/((t + 1/2)^2 + 3/4) = | t + 1/2 = u | =
= pi/2 - 2 * int (1/2 3/2) du/(u^2 + 3/4) =
= | u = 3^(1/2)/2 * v, du = 3^(1/2)/2 dv, v = 2 * u/3^(1/2) | =
= pi/2 - 2 * int (1/3^(1/2) 3^(1/2)) 3^(1/2)/2 dv/(3/4 * v^2 + 3/4) =
= pi/2 - 2 * int (1/3^(1/2) 3^(1/2)) 3^(1/2)/2 dv/(3/4 * (v^2 + 1)) =
= pi/2 - 2 * int (1/3^(1/2) 3^(1/2)) dv/(3^(1/2)/2 * (v^2 + 1)) =
= pi/2 - 2 * 2/3^(1/2) * int (1/3^(1/2) 3^(1/2)) dv/(v^2 + 1) =
= pi/2 - 4/3^(1/2) * (arctg v)_{1/3^(1/2)}^{3^(1/2)} =
= pi/2 - 4/3^(1/2) * (arctg 3^(1/2) - arctg (1/3^(1/2))) =
= pi/2 - 4/3^(1/2) * (pi/3 - pi/6) = pi/2 - 4/3^(1/2) * pi/6 =
= pi/2 - 2/3^(1/2) * 1/3 * pi = 1/18 * pi * (9 - 4 * 3^(1/2))