Question

Решить неопределенный интеграл

Answer 1

$\displaystyle \int \frac{dx}{sin^6x}=8\int\frac{dx}{(1-cos2x)^3}=8\int\frac{dt}{(1+t^2)*(1-\frac{1-t^2}{1+t^2})^3}=\\\int\frac{(t^4+2t^2+1)}{t^6}dt=\int(\frac{1}{t^2}+\frac{2}{t^4}+\frac{1}{t^6})dt=-\frac{1}{t}-\frac{2}{3t^3}-\frac{1}{5t^5}+C=\\=-\frac{1}{tg \ x}-\frac{2}{3tg^3 \ x}-\frac{1}{5tg^5 \ x}+C\\sin^2x=\frac{1-cos2x}{2}\\\\\\\\t=tg x\\arctg t =x\\dx=\frac{dt}{t^2+1}\\cos2x=\frac{1-t^2}{1+t^2}$