Вычислим интеграл:
\[ \int_{0}^{\pi/4} \frac{2dx}{\sin^2(2x+\frac{\pi}{4})} \]
Сделаем замену \( u = 2x + \frac{\pi}{4} \), тогда \( du = 2dx \).
При \( x = 0 \), \( u = \frac{\pi}{4} \).
При \( x = \frac{\pi}{4} \), \( u = 2\frac{\pi}{4} + \frac{\pi}{4} = \frac{3\pi}{4} \).
\[ = \int_{\pi/4}^{3\pi/4} \frac{du}{\sin^2(u)} = \int_{\pi/4}^{3\pi/4} \csc^2(u) du \]
\[ = [-\cot(u)]_{\pi/4}^{3\pi/4} = -(\cot(\frac{3\pi}{4}) - \cot(\frac{\pi}{4})) \]
\[ = -(-1 - 1) = -(-2) = 2 \]
Ответ: 2