Предмет: Алгебра, автор: salmantsokov2024

помогите с заданием если не сложно

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Ответы

Автор ответа: sangers1959

Объяснение:

$\displaystyle\\\int\limits^{\frac{\pi }{3} }_{\frac{\pi }{4}} {(1-cos^2x+\frac{1}{sin^2x} ) } \, dx =\int\limits^{\frac{\pi }{3} }_{\frac{\pi }{4}}dx-\int\limits^{\frac{\pi }{3} }_{\frac{\pi }{4}}cos^2xdx+\int\limits^{\frac{\pi }{3} }_{\frac{\pi }{4}}\frac{dx}{sin^2x} .\\$

$\displaystyle\\1)\ \int\limits^{\frac{\pi }{3} }_{\frac{\pi }{4}}dx=x\ |^{\frac{\pi }{4}}_{\frac{\pi }{4}} =\frac{\pi }{3}-\frac{\pi }{4} =\frac{4\pi -3\pi }{3*4}=\frac{\pi }{12}.$

$\displaystyle\\2)\ cos^2x-sin^2x=cos2x\\\\cos^2x-(1-cos^2x)=cos2x\\\\\cos^2x-1+cos^2x=cos2x\\\\ 2*cos^2x=cos2x+1\ |:2\\\\cos^2x=\frac{cos2x}{2}+\frac{1}{2} } \\\\$

$\displaystyle\\ \int\limits^{\frac{\pi }{3}} _{\frac{\pi }{4}} {cos^2x} \, dx =\int\limits^{\frac{\pi }{3}} _{\frac{\pi }{4}} \frac{cos2x}{2} +\int\limits^{\frac{\pi }{3}} _{\frac{\pi }{4}} \frac{dx}{2} =\frac{sin2x}{4}\ |^{\frac{\pi }{3}}_{\frac{\pi }{4}} +\frac{x}{2} \ |^{\frac{\pi }{3}}_{\frac{\pi }{4}} =\frac{\frac{\sqrt{3} }{2}-1 }{4} +\frac{\frac{\pi }{3}-\frac{\pi }{4} }{2}=\\\\\\$

$\displaystyle =\frac{\sqrt{3}-2 }{8} +\frac{4\pi-3\pi} {2*12} =\frac{\sqrt{3}-2 }{8} +\frac{\pi }{24}.\\$

$3)\ \displaystyle\\ \int\limits^{\frac{\pi }{3} }_{\frac{\pi }{4} } {\frac{dx}{sin^2} } =-ctgx\ |^{\frac{\pi }{3}} _{\frac{\pi }{4} }=-(\frac{\sqrt{3} }{3} -1) =1-\frac{\sqrt{3} }{3} .\ \ \ \ \Rightarrow\\$

$\displaystyle\\4)\ \frac{\pi }{12} +\frac{\sqrt{3}-2 }{8}+\frac{\pi }{24}-(1-\frac{\sqrt{3} }{3} )=\frac{\pi }{12} +\frac{\sqrt{3}-2 }{8}+\frac{\pi }{24}-1+\frac{\sqrt{3} }{3} = \\\\=\frac{2\pi +3\sqrt{3}-6 +\pi-24+8\sqrt{3} }{24} =\frac{3\pi +11\sqrt{3}-30 }{24}=\frac{\pi }{8} +\frac{11\sqrt{3} }{24} -\frac{5}{4} .$