Задача №1443
Условие
Найти интеграл \(\int\sqrt{1+\cos^2{x}}\sin{2x}\cos{2x}dx\).
Решение
\[
\int\sqrt{1+\cos^2{x}}\sin{2x}\cos{2x}dx
=\left|\begin{aligned}&u=\sqrt{1+\cos^2{x}};\;1+\cos^2{x}=u^2.\\&-2\sin{x}\cos{x}dx=2udu;\;\sin{2x}=-2udu.\\&\cos{2x}=2\cos^2{x}-1=2\cdot\left(u^2-1\right)-1=2u^2-3.\end{aligned}\right|=\\
=-\int{2u^2\left(2u^2-3\right)du}
=\int\left(-4u^4+6u^2\right)du
=-\frac{4u^5}{5}+2u^3+C
=u^3\cdot\left(-\frac{4u^2}{5}+2\right)+C=\\
=\sqrt{\left(1+\cos^2{x}\right)^3}\cdot\left(-\frac{4\cdot\left(1+\cos^2{x}\right)}{5}+2\right)
=\frac{2}{5}\cdot\sqrt{\left(1+\cos^2{x}\right)^3}\cdot\left(3-2\cos^2{x}\right)+C.
\]
Ответ:
\(\frac{2}{5}\cdot\sqrt{\left(1+\cos^2{x}\right)^3}\cdot\left(3-2\cos^2{x}\right)+C\)