32. Вычислите интеграл: \( \int_{0}^{\pi/2} \sin(2x - \frac{\pi}{4}) dx \)

Решение:

Вычислим интеграл:

\[ \int_{0}^{\pi/2} \sin(2x - \frac{\pi}{4}) dx \]

Сделаем замену \( u = 2x - \frac{\pi}{4} \), тогда \( du = 2dx \), \( dx = \frac{1}{2}du \).

При \( x = 0 \), \( u = -\frac{\pi}{4} \).

При \( x = \frac{\pi}{2} \), \( u = 2\frac{\pi}{2} - \frac{\pi}{4} = \frac{3\pi}{4} \).

\[ = \int_{-\pi/4}^{3\pi/4} \sin(u) \frac{1}{2} du = \frac{1}{2} \int_{-\pi/4}^{3\pi/4} \sin(u) du \]

\[ = \frac{1}{2} [-\cos(u)]_{-\pi/4}^{3\pi/4} = -\frac{1}{2} [\cos(u)]_{-\pi/4}^{3\pi/4} \]

\[ = -\frac{1}{2} (\cos(\frac{3\pi}{4}) - \cos(-\frac{\pi}{4})) \]

\[ = -\frac{1}{2} (-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}) = -\frac{1}{2} (-\sqrt{2}) = \frac{\sqrt{2}}{2} \]

Ответ: \(\frac{\sqrt{2}}{2}\)