Помогите, пожалуйста, найти интеграл int dx/sin^3 x
I = int dx/sin^3 x = int 1/sin x * 1/sin^2 x dx = int 1/sin x d(-ctg x) =
= -int 1/sin x d(ctg x) = -1/sin x * ctg x + int ctg x d(1/sin x) =
= -1/sin x * cos x/sin x + int cos x/sin x * (-cos x)/sin^2 x dx =
= -cos x/sin^2 x - int cos^2 x/sin^3 x dx =
= -cos x/sin^2 x - int (1 - sin^2 x)/sin^3 x dx =
= -cos x/sin^2 x - int dx/sin^3 x + int dx/sin x.
int dx/sin x = int dx/(2 * sin (x/2) * cos (x/2)) =
= 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)) = | t = tg (x/2) | =
= int 1/t dt = ln |t| + C = | t = tg (x/2) | = ln |tg (x/2)| + C
Получаем, что
I = -cos x/sin^2 x - I + ln |tg (x/2)| + C
2 * I = -cos x/sin^2 x + ln |tg (x/2)| + C
I = -1/2 * cos x/sin^2 x + 1/2 * ln |tg (x/2)| + C
int dx/sin^3 x = -1/2 * cos x/sin^2 x + 1/2 * ln |tg (x/2)| + C
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