Question

Решите пожалуйста, даю 10 баллов

Answer 1

Ответ:

$\frac{\sqrt{2} }{2}$

Пошаговое объяснение:

$\int\limits^\frac{\pi }{2} _\frac{\pi }{4} {sin(2x-\frac{\pi }{4} )} \, dx$

1) $\int\limits {sin(2x-\frac{\pi }{4} )} \, dx$              

                                           $t=2x-\frac{\pi }{4} \\t'=(2x)'-(\frac{\pi }{4} )'\\t'=2-0\\dt=2dx$

                                            $(t'=\frac{dt}{dx})$

$\int\limits {sin(2x-\frac{\pi }{4} )} \, dx = \int\limits {sin(t)} \, \frac{dt}{2} = \int\limits {\frac{sin(t)}{2} } \, dt=\frac{1}{2} \int\limits {sin(t) } \, dt=\frac{1}{2} *(-cos(t))$

$\frac{1}{2} *(-cos(2x-\frac{\pi }{4} )=-\frac{\sqrt{2}cos(2x)+\sqrt{2}sin(2x) }{4}$

2)  

$-\frac{\sqrt{2}cos(2x)+\sqrt{2}sin(2x) }{4}=-\frac{\sqrt{2}cos(2*\frac{\pi }{2} )+\sqrt{2}sin(2*\frac{\pi }{2})}{4} -\frac{\sqrt{2}cos(2*\frac{\pi }{4} )+\sqrt{2}sin(2*\frac{\pi }{4})}{4} =\\-\frac{\sqrt{2}cos(\pi )+\sqrt{2}sin(\pi )}{4}-(-\frac{\sqrt{2}cos(\frac{\pi }{2} )+\sqrt{2}sin(\frac{\pi }{2})}{4})=-\frac{\sqrt{2}*(-1)+\sqrt{2}*0}{4}-(-\frac{\sqrt{2}*0+\sqrt{2}*1 }{4} )=-\frac{-\sqrt{2} }{4}-(-\frac{\sqrt{2}}{4})= \frac{\sqrt{2} }{4}+\frac{\sqrt{2}}{4}=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2}$