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∫₀ sinxcosx dx=∫₀ (1/2* sin2x) dx=1/2*( -1/2cos2x) |₀= -1/4(cosπ-cos0)= -1/4(-1-1)=1/2

Воспользовались формулой sin2x=2sinxcosx

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Nik133

$\int\limits_0^{\pi/2}{sinx*cosx}dx=\frac{1}{2}\int\limits_0^{\pi/2}{2sinx*cosx}dx= \frac{1}{2}\int\limits_0^{\pi/2}{sin2x}dx= \\ \\ =\frac{1}{2}*(-\frac{1}{2}*cos2x)[_0^{\pi/2}= -\frac{1}{4}*cos2x[_0^{\pi/2}= \\ \\ =-\frac{1}{4}*cos{(2*\frac{\pi}{2})}-(-\frac{1}{4}*cos0)= -\frac{1}{4}*(-1)+\frac{1}{4}*1= \\ \\ =\frac{1}{4}+\frac{1}{4}=\frac{1}{2}=0,5$

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