Question

Решить неопределенный интеграл

Answer 1

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