интеграл cosec^5x*dx нужно решить за час!ПОМОГИТЕ РЕШИТЬ ПОЖАЛУЙСТА!

I = int cosec^5 x dx = int 1/sin^5 x dx =
= int 1/sin^3 x * 1/sin^2 x dx = int 1/sin^3 x d(-ctg x) =
= -ctg x/sin^3 x + int ctg x d(1/sin^3 x) =
= -(cos x/sin x)/sin^3 x + int cos x/sin x * (-3) * cos x/sin^4 x dx =
= -cos x/sin^4 x - 3 * int cos^2 x/sin^5 x dx =
= -cos x/sin^4 x - 3 * int (1 - sin^2 x)/sin^5 x dx =
= -cos x/sin^4 x - 3 * int 1/sin^5 x dx + 3 * int 1/sin^3 x dx
Получаем, что
I = -cos x/sin^4 x - 3 * I + 3 * int 1/sin^3 x dx
4 * I = -cos x/sin^4 x + 3 * int 1/sin^3 x dx
I = -1/4 * cos x/sin^4 x + 3/4 * int 1/sin^3 x dx

int 1/sin^3 x dx = int 1/sin x d(-ctg x) =
= -ctg x/sin x + int ctg x d(1/sin x) =
= -ctg x/sin x + int cos x/sin x * (-cos x/sin^2 x) dx =
= -(cos x/sin x)/sin x - int (1 - sin^2 x)/sin^3 x dx =
= -cos x/sin^2 x - int 1/sin^3 x dx + int dx/sin x
int 1/sin^3 x = -1/2 * cos x/sin^2 x + 1/2 * int 1/sin x dx

int 1/sin x dx = int 1/(2 * sin (x/2) * cos (x/2)) dx =
= int cos (x/2)/(2 * sin (x/2) * cos^2 (x/2)) dx =
= int cos (x/2)/sin (x/2) d(tg (x/2)) = int 1/tg (x/2) d(tg (x/2)) =
= ln |tg (x/2)| + C
Тогда
int 1/sin^3 x = -1/2 * cos x/sin^2 x + 1/2 * ln |tg (x/2)| + C
I = -1/4 * cos x/sin^4 x + 3/4 * int 1/sin^3 x dx =
= -1/4 * cos x/sin^4 x - 3/8 * cos x/sin^2 x + 3/8 * ln |tg (x/2)| + C
Вот такой ответ кажется получается.