Question

Сделайте пожалуйста интеграл

Answer 1

$\displaystyle \int\frac{dx}{sin^2xcos^4x}=8\int\frac{dx}{(1-cos2x)(1+cos2x)^2}=\\=8\int\frac{dt}{(1+t^2)(1-\frac{1-t^2}{1+t^2})(1+\frac{1-t^2}{1+t^2})^2}=\int\frac{t^4+2t^2+1}{t^2}dt=\\=\int(t^2+2+\frac{1}{t^2})dt=\frac{t^3}{3}+2t-\frac{1}{t}+C=\frac{tg^3x}{3}+2tgx-\frac{1}{tgx}+C\\\\\\sin^2x=\frac{1-cos2x}{2};cos^2x=\frac{1+cos2x}{2}\\t=tgx\\arctg\ t=x\\dx=\frac{dt}{1+t^2}\\cos2x=\frac{1-t^2}{1+t^2}$