Здравствуйте.
Помогите, пожалуйста, найти интеграл
int sin^6 (x/3) dx
Заранее спасибо.

int sin^6 (x/3) dx = | t = x/3, x = 3 * t, dx = 3 dt | =
= int 3 * sin^6 t dt = 3 * int sin^6 t dt = 3 * int (sin^2 t)^3 dt =
= 3 * int ((1 - cos 2t)/2)^3 dt = 3 * int (1 - cos 2t)^3/8 dt =
= 3/8 * int (1 - cos 2t)^3 dt = 3/8 * int (1 - 3 * cos 2t + 3 * cos^2 2t - cos^3 2t) dt =
= 3/8 * int dt - 9/8 * int cos 2t dt + 9/8 * int cos^2 2t dt - 3/8 * int cos^3 2t dt =
= 3/8 * t - 9/8 * 1/2 * sin 2t + 9/8 * int (1 + cos 4t)/2 dt -
- 3/8 * int cos^2 2t * cos 2t dt =
= 3/8 * t - 9/16 * sin 2t + 9/16 * int (1 + cos 4t) dt - 3/8 * int cos^2 2t d(1/2 * sin 2t) =
= 3/8 * t - 9/16 * sin 2t + 9/16 * (t + 1/4 * sin 4t) - 3/16 * int (1 - sin^2 2t) d(sin 2t) =
= | sin 2t = u | =
= 3/8 * t - 9/16 * sin 2t + 9/16 * t + 9/64 * sin 4t - 3/16 * int (1 - u^2) du =
= 3/8 * t - 9/16 * sin 2t + 9/16 * t + 9/64 * sin 4t - 3/16 * (u - 1/3 * u^3) + C =
= 3/8 * t - 9/16 * sin 2t + 9/16 * t + 9/64 * sin 4t - 3/16 * u + 1/16 * u^3 + C =
= | u = sin 2t | =
= 3/8 * t - 9/16 * sin 2t + 9/16 * t + 9/64 * sin 4t - 3/16 * sin 2t + 1/16 * sin^3 2t + C =
= 6/16 * t + 9/16 * t - 12/16 * sin 2t + 9/64 * sin 4t + 1/16 * sin^3 2t + C =
= 15/16 * t - 3/4 * sin 2t + 9/64 * sin 4t + 1/16 * sin^3 2t + C = | t = x/3 | =
= 15/16 * x/3 - 3/4 * sin (2x/3) + 9/64 * sin (4x/3) + 1/16 * sin^3 (2x/3) + C =
= 5/16 * x - 3/4 * sin (2x/3) + 9/64 * sin (4x/3) + 1/16 * sin^3 (2x/3) + C

Cпасибо!