ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 394

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

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

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

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

\[= \frac{1}{2} \bullet \frac{\sqrt{2}}{2} - \frac{1}{2} \bullet \frac{\sqrt{2}}{2} = 0.\]

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

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

\[= \frac{1}{3}\sin\left( \frac{\pi}{3} - \frac{\pi}{3} \right) - \frac{1}{3}\sin\left( 0 - \frac{\pi}{3} \right) =\]

\[= \frac{1}{3}\sin 0 + \frac{1}{3}\sin\frac{\pi}{3} =\]

\[= \frac{1}{3} \bullet 0 + \frac{1}{3} \bullet \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{6}.\]

\[3)\ \int_{1}^{3}{3\sin(3x - 6)\text{dx}} =\]

\[= \left. \ \left( - 3 \bullet \frac{1}{3}\cos(3x - 6) \right) \right|_{1}^{3} =\]

\[= \left. \ - \cos(3x - 6) \right|_{1}^{3} =\]

\[= - \cos(3 \bullet 3 - 6) + \cos(3 \bullet 1 - 6) =\]

\[= - \cos 3 + \cos( - 3) = 0.\]

\[4)\ \int_{0}^{3}{8\cos(4x - 12)\text{dx}} =\]

\[= \left. \ \left( 8 \bullet \frac{1}{4}\sin(4x - 12) \right) \right|_{0}^{3} =\]

\[= \left. \ 2\sin(4x - 12) \right|_{0}^{3} =\]

\[= 2\sin(4 \bullet 3 - 12) - 2\sin(4 \bullet 0 - 12) =\]

\[= 2\sin 0 - 2\sin( - 12) = 2\sin 12.\]