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Решите задачу:

$\\\int \limits_{\frac{\pi}{18}}^{\frac{\pi}{4}} {\cos 4x}\, dx=(*)\\ t=4x\\ dt=4\, dx\\ \int \limits_{\frac{\pi}{18}}^{\frac{\pi}{4}} \frac{1}{4}{\cos t}\, dt=\\ \frac{1}{4}\int \limits_{\frac{\pi}{18}}^{\frac{\pi}{4}} {\cos t}\, dt=\\ \frac{1}{4}\Big[\sin t\Big]_{\frac{\pi}{18}}^{\frac{\pi}{4}}=\\ (*)=\frac{1}{4}\Big[\sin 4x\Big]_{\frac{\pi}{18}}^{\frac{\pi}{4}}=\\ \frac{1}{4}(\sin (4\cdot\frac{\pi}{4})-\sin (4\cdot\frac{\pi}{18}))=\\\frac{1}{4}(\sin \pi-\sin \frac{2\pi}{9})=\\\frac{1}{4}(0-\sin \frac{2\pi}{9})=\\-\frac{1}{4}\sin \frac{2\pi}{9}\approx-0,16\\$