Задача №1620

\[ \int\frac{dx}{1-\sin^4{x}} =\int\frac{dx}{\left(1-\sin^2{x}\right)\left(1+\sin^2{x}\right)} =\int\frac{dx}{\cos^2{x}\left(2-\cos^2{x}\right)}=\\ =\int\frac{d(\tg{x})}{2-\frac{1}{\tg^2{x}+1}} =\left[t=\tg{x}\right] =\int\left(\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2t^2+1}\right)dt =\frac{\tg{x}}{2}+\frac{1}{2\sqrt{2}}\arctg\left(\sqrt{2}\tg{x}\right)+C \]