Question

Помогите вычислить интеграл!

Answer 1

$\int\dfrac{1}{\sin^2\dfrac{x}{2}}dx\\\\\\t=\dfrac{x}{2}\\\\\\\int\dfrac{2}{\sin^2t}dt\\\\\\2\cdot\int\dfrac{1}{\sin^2t}dt\\\\\\2\cdot(-\cot{t})\\\\\\2\cdot\bigg(-\cot\dfrac{x}{2}\bigg)\\\\\\-2\cot\dfrac{x}{2}\\\\\\-2\cot\dfrac{x}{2}+C$

Answer 2

Решение:

$\int \frac{1}{sin^2\frac{x}{2} } dx\\\\t = \frac{x}{2}\\\\dt = d(\frac{x}{2}) = (\frac{x}{2}) `dx = \frac{dx}{2}\\\\dt = \frac{dx}{2}\\\\dx= 2dt\\\\\int \frac{1}{sin^2t } 2dt = 2\int \frac{1}{sin^2t } dt$

Мы знаем, что:

$\int \frac{1}{sin^2x} dx = -ctg(x)+C$

Получаем:

$2\int \frac{1}{sin^2t } dt = 2 *(-ctg(t)) = -2ctg(t) = -2ctg\frac{x}{2} + C$