Задача №1613

\[ \int\frac{dx}{\sin^4{x}\cos^4{x}} =\int\frac{16dx}{\sin^4{2x}} =8\cdot\int\frac{d(\ctg{2x})}{\sin^2{2x}} =-8\cdot\int\left(\ctg^2{2x}+1\right)d(\ctg{2x}) =-\frac{8\ctg^3{2x}}{3}-8\ctg{2x}+C \]

\[ \int\frac{dx}{\sin^4{x}\cos^4{x}} =\int\left(\frac{1}{\sin^2{x}}\right)^2\cdot\frac{1}{\cos^2{x}}\cdot\frac{dx}{\cos^2{x}}=\\ =\int\left(\frac{1}{\tg^2{x}}+1\right)^2\cdot\left(1+\tg^2{x}\right)^2 d(\tg{x}) =[u=\tg{x}] =\int\left(\frac{1}{u^4}+\frac{3}{u^2}+u^2+3\right)du =\ldots \]