1) int (2 * x + 9)/(x^2 + 5 * x + 6) dx, 2) int sin^3 x * cos^5 x dx Опции

[Xiu Xiu]

Помогите, пожалуйста, найти два интеграла:
1) int (2 * x + 9)/(x^2 + 5 * x + 6) dx
2) int sin^3 x * cos^5 x dx

[tig81]

1) int (2 * x + 9)/(x^2 + 5 * x + 6) dx = int (2 * x + 9)/((x + 2) * (x + 3)) dx
Разложим подынтегральное выражение на простейшие дроби:
(2 * x + 9)/((x + 2) * (x + 3)) = A/(x + 2) + B/(x + 3)
Умножим обе части равенства на (x + 2) * (x + 3):
2 * x + 9 = A * (x + 3) + B * (x + 2)
x = -2 => 2 * (-2) + 9 = A * (-2 + 3) => A = 5
x = -3 => 2 * (-3) + 9 = B * (-3 + 2) => B = -3
Тогда
int (2 * x + 9)/((x + 2) * (x + 3)) dx = int (5/(x + 2) - 3/(x + 3)) dx =
= 5 * int dx/(x + 2) - 3 * int dx/(x + 3) = 5 * ln |x + 2| - 3 * ln |x + 3| + C
2) int sin^3 x * cos^5 x dx = int sin^2 x * cos^5 x * sin x dx =
= int sin^2 x * cos^5 x d(-cos x) = -int sin^2 x * cos^5 x d(cos x) =
= -int (1 - cos^2 x) * cos^5 x d(cos x) = | cos x = t | =
= -int (1 - t^2) * t^5 dt = -int (t^5 - t^7) dt = -(1/6 * t^6 - 1/8 * t^8) + C =
= 1/8 * t^8 - 1/6 * t^6 + C = | t = cos x | = 1/8 * cos^8 x - 1/6 * cos^6 x + C