Задача №1646
Условие
Найти интеграл \(\int\frac{dx}{\sin^4{x}+\cos^4{x}}\).
Решение
\[
\int\frac{dx}{\sin^4{x}+\cos^4{x}}
=\int\frac{dx}{\left(\sin^2{x}+\cos^2{x}\right)^2-2\sin^2{x}\cos^2{x}}
=\int\frac{dx}{1-\frac{\sin^2{2x}}{2}}=\\
=\int\frac{2dx}{2-\sin^2{2x}}
=[u=2x]
=\int\frac{du}{2-\sin^2{u}}
=\int\frac{du}{1+\cos^2{u}}
=\int\frac{\frac{du}{\cos^2{u}}}{\frac{1}{\cos^2{u}}+1}=\\
=\int\frac{d(\tg{u})}{\tg^2{u}+2}
=\frac{1}{\sqrt{2}}\arctg\frac{\tg{u}}{\sqrt{2}}+C
=\frac{1}{\sqrt{2}}\arctg\frac{\tg{2x}}{\sqrt{2}}+C
\]
Ответ:
\(\frac{1}{\sqrt{2}}\arctg\frac{\tg{2x}}{\sqrt{2}}+C\)