Răspuns:
Explicație pas cu pas:
[tex]\int\limits^{\frac{\pi }{6} }_{\frac{\pi }{18} } {cos3x} \, dx=\frac{1}{3}*sin3x|_{\frac{\pi }{18} }^{\frac{\pi }{6} } =\frac{1}{3}*(sin(3*\frac{\pi }{6} )-sin(3*\frac{\pi }{18} )=\frac{1}{3}*(sin\frac{\pi }{2}-sin\frac{\pi }{6} )=\frac{1}{3} *(1-\frac{1}{2})=\frac{1}{3} *\frac{1}{2}= \frac{1}{6}.[/tex]