первообразная [tex] \frac{1}{sin^25x} [/tex] равна [tex]-ctg5x?[/tex]. Created by karavanov1. algebra-ru

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July 2022 1 7 Report

первообразная [tex] \frac{1}{sin^25x} [/tex] равна [tex]-ctg5x?[/tex]

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laymlaym2 $\int\frac{1}{sin^2(5x)}dx=[d(5x)=5dx\rightarrow dx=\frac{d(5x)}{5}]=\int\frac{1}{sin^2(5x)}*\frac{d(5x)}{5}=\\=\frac{1}{5}\int\frac{d(5x)}{sin^2(5x)}=\frac{1}{5}*(-ctg(5x))=-\frac{ctg(5x)}{5}$

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zaputalsya pomogite s trigonometriej vot cos80 kakoe iz preobrazovanij schitaet

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mozhno li najti pervoobraznuyu bez pomoshi tablicy tolko nachal prohodit temu per

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