Question

Помогите решить интеграл!!!!

Answer 1

Ответ:

$\frac{\sqrt{2} }{4} + \frac{33\pi }{8}$

Объяснение:

$\int\limits^{33\pi /8}_{-33\pi /8} {cos^2(x-2\pi )} \, dx = \int\limits^{33\pi /8}_{-33\pi /8} {\frac{cos(2x-4\pi )+1}{2} } \, dx =\frac{1}{4}sin(2x-4\pi ) + \frac{x}{2} |^{33\pi /8}_{-33\pi /8}=$

$=\frac{1}{4}(sin(\frac{33\pi }{4}-\frac{16\pi }{4}) - sin(-\frac{33\pi }{4}-\frac{16\pi }{4})) +\frac{33\pi }{16} - (-\frac{33\pi }{16} ) =$

$=\frac{1}{4} (sin(\frac{17\pi }{4} )) + sin(\frac{49\pi }{4} )+\frac{33\pi }{8} =\frac{1}{4}(sin(\frac{\pi }{4} )+sin(\frac{\pi }{4} ))+\frac{33\pi }{8}=$

$=\frac{1}{4}(\frac{\sqrt{2} }{2} + \frac{\sqrt{2} }{2}) + \frac{33\pi }{8} =\frac{\sqrt{2} }{4} + \frac{33\pi }{8}$