Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 375

\[1)\ \int_{0}^{\frac{\pi}{2}}{\sin{2x}\text{dx}} = \left. \ - \frac{1}{2}\cos{2x} \right|_{0}^{\frac{\pi}{2}} =\]

\[= - \frac{1}{2}\cos\pi + \frac{1}{2}\cos 0 =\]

\[= - \frac{1}{2} \bullet ( - 1) + \frac{1}{2} \bullet 1 = \frac{1}{2} + \frac{1}{2} = 1.\]

\[2)\ \int_{\frac{\pi}{4}}^{\pi}{\cos\left( 3x - \frac{\pi}{4} \right)\text{dx}} =\]

\[= \left. \ \frac{1}{3}\sin\left( 3x - \frac{\pi}{4} \right) \right|_{\frac{\pi}{4}}^{\pi} =\]

\[= \frac{1}{3}\sin\left( 3\pi - \frac{\pi}{4} \right) - \frac{1}{3}\sin\left( \frac{3\pi}{4} - \frac{\pi}{4} \right) =\]

\[= \frac{1}{3}\sin\frac{\pi}{4} - \frac{1}{3}\sin\frac{\pi}{2} =\]

\[= \frac{1}{3} \bullet \frac{\sqrt{2}}{2} - \frac{1}{3} \bullet 1 = \frac{\sqrt{2}}{6} - \frac{1}{3}.\ \]

\[3)\ \int_{0}^{\frac{\pi}{4}}{\cos^{2}{4x}\text{dx}} =\]

\[= \int_{0}^{\frac{\pi}{4}}{\frac{1 + \cos{8x}}{2}\text{dx}} =\]

\[= \int_{0}^{\frac{\pi}{4}}{\left( \frac{1}{2} + \frac{1}{2}\cos{8x} \right)\text{dx}} =\]

\[= \left. \ \left( \frac{1}{2}x + \frac{1}{2} \bullet \frac{1}{8}\sin{8x} \right) \right|_{0}^{\frac{\pi}{4}} =\]

\[= \left. \ \left( \frac{1}{2}x + \frac{1}{16}\sin{8x} \right) \right|_{0}^{\frac{\pi}{4}} =\]

\[= \frac{\pi}{8} + \frac{1}{16} \bullet 0 - \frac{1}{16} \bullet 0 = \frac{\pi}{8}.\]

\[4)\ \int_{0}^{\frac{\pi}{3}}{\sin^{2}\left( x - \frac{\pi}{3} \right)\text{dx}} =\]

\[= \int_{0}^{\frac{\pi}{3}}{\frac{1 - \cos\left( 2x - \frac{2\pi}{3} \right)}{2}\text{dx}} =\]

\[= \int_{0}^{\frac{\pi}{3}}{\left( \frac{1}{2} - \frac{1}{2}\cos\left( 2x - \frac{2\pi}{3} \right) \right)\text{dx}} =\]

\[= \left. \ \left( \frac{1}{2}x - \frac{1}{2} \bullet \frac{1}{2}\sin\left( 2x - \frac{2\pi}{3} \right) \right) \right|_{0}^{\frac{\pi}{3}} =\]

\[= \frac{\pi}{6} - \frac{1}{4}\sin 0 + \frac{1}{4} \bullet \left( - \frac{\sqrt{3}}{2} \right) =\]

\[= \frac{\pi}{6} - \frac{\sqrt{3}}{8} = \frac{4\pi - 3\sqrt{3}}{24}.\]