Помогите, пожалуйста, вычислить интеграл:
int (0 1) (4 - x^2)^(1/2) dx

Сначала вычислим неопределенный интеграл int (4 - x^2)^(1/2) dx.
I = int (4 - x^2)^(1/2) dx = x * (4 - x^2)^(1/2) - int x d((4 - x^2)^(1/2)) =
= x * (4 - x^2)^(1/2) - int x * (-2x) * 1/2 * 1/(4 - x^2)^(1/2) dx =
= x * (4 - x^2)^(1/2) - int (-x^2)/(4 - x^2)^(1/2) dx =
= x * (4 - x^2)^(1/2) - int ((4 - x^2) - 4)/(4 - x^2)^(1/2) dx =
= x * (4 - x^2)^(1/2) - int (4 - x^2)/(4 - x^2)^(1/2) dx + int 4/(4 - x^2)^(1/2) dx =
= x * (4 - x^2)^(1/2) - int (4 - x^2)^(1/2) dx + 4 * int 1/(4 - x^2)^(1/2) dx =
= x * (4 - x^2)^(1/2) - I + 4 * arcsin (x/2)
I = x * (4 - x^2)^(1/2) - I + 4 * arcsin (x/2)
2 * I = x * (4 - x^2)^(1/2) + 4 * arcsin (x/2) + C
Получаем, что
I = x/2 * (4 - x^2)^(1/2) + 2 * arcsin (x/2) + C
Тогда
int (0 1) (4 - x^2)^(1/2) dx = (x/2 * (4 - x^2)^(1/2) + 2 * arcsin (x/2))_{0}^{1} =
= (1/2 * (4 - 1^2)^(1/2) + 2 * arcsin (1/2)) - (0/2 * (4 - 0^2)^(1/2) + 2 * arcsin 0) =
= 1/2 * 3^(1/2) + 2 * pi/6 = 3^(1/2)/2 + pi/3