Здравствуйте, помогите, пожалуйста, решить задачу:
найти площадь поверхности, образованной вращением кривой y = tg x от ее точки (0;0) до точки (pi/4;1) вокруг оси Ox.

y = tg x, A(0;0) B(pi/4;1)
Используем формулу для вычисления площади поверхности вращения кривой
y = y(x) вокруг оси Ох.
S = 2 * pi * int (a b ) |y| * (1 + (y')^2)^(1/2) dx.
Получаем, что
S = 2 * pi * int (0 pi/4) |tg x| * (1 + ((tg x)')^2)^(1/2) dx =
= 2 * pi * int (0 pi/4) tg x * (1 + (1/cos^2 x)^2)^(1/2) dx =
= 2 * pi * int (0 pi/4) tg x * (1 + 1/cos^4 x)^(1/2) dx =
= 2 * pi * int (0 pi/4) sin x/cos x * ((cos^4 x + 1)/cos^4 x)^(1/2) dx =
= 2 * pi * int (0 pi/4) sin x/cos x * (cos^4 x + 1)^(1/2)/cos^2 x dx =
= 2 * pi * int (0 pi/4) (cos^4 x + 1)^(1/2)/cos^3 x * sin x dx =
= 2 * pi * int (0 pi/4) (cos^4 x + 1)^(1/2)/cos^3 x d(-cos x) =
= -2 * pi * int (0 pi/4) (cos^4 x + 1)^(1/2)/cos^3 x d(cos x) = | t = cos x | =
= -2 * pi * int (1 2^(1/2)/2) (t^4 + 1)^(1/2)/t^3 dt =
= 2 * pi * int (2^(1/2)/2 1) (t^4 + 1)^(1/2) * t^(-3) dt =
= | t^(-4) + 1 = z^2, z = (1 + 1/t^4)^(1/2), t^4 = 1/(z^2 - 1), t = (z^2 - 1)^(-1/4),
dt = -1/4 * (z^2 - 1)^(-5/4) * 2 * z | =
= 2 * pi * int (5^(1/2) 2^(1/2)) (1/(z^2 - 1) + 1)^(1/2) * (z^2 - 1)^(3/4) *
* (-1/4) * (z^2 - 1)^(-5/4) * 2 * z dz =
= 2 * pi * int (5^(1/2) 2^(1/2)) (z^2/(z^2 - 1))^(1/2) * (z^2 - 1)^(-1/2) *
* (-1/2) * z dz =
= -pi * int (5^(1/2) 2^(1/2)) z^2/(z^2 - 1)^(1/2) * 1/(z^2 - 1)^(1/2) dz =
= pi * int (2^(1/2) 5^(1/2)) z^2/(z^2 - 1) dz =
= pi * int (2^(1/2) 5^(1/2)) (z^2 - 1 + 1)/(z^2 - 1) dz =
= pi * int (2^(1/2) 5^(1/2)) dz + pi * int (2^(1/2) 5^(1/2)) 1/(z^2 - 1) dz =
= pi * (z)_{2^(1/2)}^{5^(1/2)} + pi * int (2^(1/2) 5^(1/2)) 1/((z - 1) * (z + 1)) dz =
= pi * (5^(1/2) - 2^(1/2)) + 1/2 * pi * int (2^(1/2) 5^(1/2)) 2/((z - 1) * (z + 1)) dz =
= pi * (5^(1/2) - 2^(1/2)) +
+ 1/2 * pi * int (2^(1/2) 5^(1/2)) ((z + 1) - (z - 1))/((z - 1) * (z + 1)) dz =
= pi * (5^(1/2) - 2^(1/2)) + 1/2 * pi * int (2^(1/2) 5^(1/2)) dz/(z - 1) -
- 1/2 * pi * int (2^(1/2) 5^(1/2)) dz/(z + 1) =
= pi * (5^(1/2) - 2^(1/2)) + 1/2 * pi * (ln |z - 1|)_{2^(1/2)}^{5^(1/2)} -
- 1/2 * pi * (ln |z + 1|)_{2^(1/2)}^{5^(1/2)} =
= pi * (5^(1/2) - 2^(1/2)) + 1/2 * pi * (ln |5^(1/2) - 1| - ln |2^(1/2) - 1|) -
- 1/2 * pi * (ln |5^(1/2) + 1| - ln |2^(1/2) + 1|) =
= pi * (5^(1/2) - 2^(1/2)) +
+ 1/2 * pi * (ln |5^(1/2) - 1| - ln |2^(1/2) - 1| - ln |5^(1/2) + 1| + ln |2^(1/2) + 1|) =
= pi * (5^(1/2) - 2^(1/2)) +
+ 1/2 * pi * ln (((5^(1/2) - 1) * (2^(1/2) + 1))/((2^(1/2) - 1) * (5^(1/2) + 1)))
Преобразуем выражение под логарифмом:
(5^(1/2) - 1)/(5^(1/2) + 1) = (5^(1/2) - 1)^2/((5^(1/2) + 1) * (5^(1/2) - 1)) =
= (5^(1/2) - 1)^2/4
(2^(1/2) + 1)/(2^(1/2) - 1) = (2^(1/2) + 1)^2/((2^(1/2) - 1) * (2^(1/2) + 1)) =
= (2^(1/2) + 1)^2
Тогда
S = pi * (5^(1/2) - 2^(1/2)) + 1/2 * pi * ln (((5^(1/2) - 1)^2 * (2^(1/2) + 1)^2)/4) =
= pi * (5^(1/2) - 2^(1/2)) + 1/2 * pi * ln ((5^(1/2) - 1) * (2^(1/2) + 1))/2)^2 =
= pi * (5^(1/2) - 2^(1/2)) + pi * ln ((5^(1/2) - 1) * (2^(1/2) + 1))/2)
Ответ: S = pi * (5^(1/2) - 2^(1/2)) + pi * ln ((5^(1/2) - 1) * (2^(1/2) + 1))/2)

Огромное спасибо!!!