Ответ:

[tex]\boldsymbol{\boxed{\int\limits^{\pi }_{-\pi } {\bigg (4 \cos 4x + \frac{1}{3} \sin \frac{x}{3} \bigg )} \, dx =0}}[/tex]

Примечание:

По таблице интегралов:

[tex]\boxed{\int {\sin x} \, dx =-\cos x+C }[/tex]

[tex]\boxed{\int {\cos x} \, dx =\sin x+C }[/tex]

По свойствам интегралов:

[tex]\boxed{ \displaystyle \int\limits^a_b \sum\limits_{i=1}^n {C_{i}f_{i}(x)} \, dx = \sum\limits_{i=1}^nC_{i} \int\limits^a_b {f_{i}(x)} \, dx}[/tex]

Объяснение:

[tex]\displaystyle \int\limits^{\pi }_{-\pi } {\bigg (4 \cos 4x + \frac{1}{3} \sin \frac{x}{3} \bigg )} \, dx = \int\limits^{\pi }_{-\pi } {4 \cos 4x } \, dx +\int\limits^{\pi }_{-\pi } {\frac{1}{3} \sin \frac{x}{3} } \, dx =[/tex]

[tex]\displaystyle = \int\limits^{\pi }_{-\pi } { \cos 4x } \, d(4x) +\int\limits^{\pi }_{-\pi } {\sin \frac{x}{3} } \, d\bigg (\frac{x}{3} \bigg) = \sin 4x \bigg |_{-\pi }^{\pi } - \cos \frac{x}{3} \bigg |_{-\pi }^{\pi } =[/tex]

[tex]= \sin 4\pi - \sin (-4\pi ) - \bigg ( \cos \bigg ( \dfrac{\pi }{3} \bigg ) - \cos \bigg ( -\dfrac{\pi }{3} \bigg ) \bigg) =[/tex]

[tex]= \sin 4\pi + \sin 4\pi - \bigg ( \cos \bigg ( \dfrac{\pi }{3} \bigg ) - \cos \bigg ( \dfrac{\pi }{3} \bigg ) \bigg) =2\sin 4\pi =2\sin(0+ 2 \cdot 2\pi) = 2 \sin 0 =[/tex]

[tex]= 2 \cdot 0 = 0[/tex]