tex]...

Вовина79 @Вовина79

July 2022 1 5 Report

Найдите первообразную для функции:[tex]f(x)=\frac{sin(\frac{\pi }{6}-2x )}{cos^{3} (\frac{\pi }{3}+2x) }[/tex]

Answers & Comments

teledima00

$\frac{\pi}{3} = \frac{\pi}{2}-\frac{\pi}{6}\\\\\int\limits \frac{sin(\frac{\pi}{6}-2x)}{cos^3(\frac{\pi}{3}+2x)} \, dx = \int\limits \frac{sin(\frac{\pi}{6}-2x)}{cos^3(\frac{\pi}{2}-\frac{\pi}{6}+2x)} \, dx = \int\limits\frac{sin(\frac{\pi}{6}-2x)}{cos^3(\frac{\pi}{2}-(\frac{\pi}{6}-2x))} \, dx =\\\\= \int\limits\frac{sin(\frac{\pi}{6}-2x)}{sin^3(\frac{\pi}{6}-2x)} \, dx = \int\limits\frac{dx}{sin^2(\frac{\pi}{6}-2x)} = -\frac{1}{2}\int\limits\frac{d(\frac{\pi}{6}-2x)}{sin^2(\frac{\pi}{6}-2x)} =$

$= \frac{1}{2}ctg(\frac{\pi}{6}-2x) + C$

Ответ: $\frac{1}{2}ctg(\frac{\pi}{6}-2x) + C$

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