Question

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Answer 1

$y = 2x + \frac{\pi}{6}$

При $x = 0$ имеем $y = \frac{\pi}{6}$.

При $x = \frac{\pi}{12}$ имеем $y = 2\cdot\frac{\pi}{12} + \frac{\pi}{6} =$

$= \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$

$\mathrm{d}y = \mathrm{d}(2x+ \frac{\pi}{6}) = 2\mathrm{d}x$

$\mathrm{d}x = \frac{\mathrm{d}y}{2}$

$\int\limits_0^{\frac{\pi}{12}} \sin\left( 2x + \frac{\pi}{6} \right)\;\mathrm{d}x =$

$= \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin(y)\cdot\frac{\mathrm{d}y}{2} =$

$= \frac{1}{2}\cdot\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin(y)\mathrm{d}y =$

$= \frac{1}{2}\cdot( -\cos(y))|\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} =$

$= \frac{1}{2}\cdot\left( -\cos(\frac{\pi}{3}) - (-\cos(\frac{\pi}{6})) \right) =$

$= \frac{1}{2}\cdot\left( -\frac{1}{2} + \frac{\sqrt{3}}{2} \right) =$

$= \frac{\sqrt{3} - 1}{4}$