Задача №1646

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