Ответ:

[tex]\displaystyle \int\limits^\pi _0 \left ( \frac{1}{3}\cos \frac{x}{3} +4 \sin 4x \right ) \, dx=\frac{\sqrt{3} }{2}[/tex]

Объяснение:

[tex]\displaystyle \int\limits^\pi _0 \left ( \frac{1}{3}\cos \frac{x}{3} +4 \sin 4x \right ) \, dx[/tex]

Найдем неопределенный интеграл

[tex]\bullet ~~ \displaystyle \boldsymbol{\int\limits \left (f(x) \pm g(x) \right )\, dx = \int\limits{f(x)} \, dx \pm \int\limits{g(x)} \, dx } }[/tex]

[tex]\bullet ~~ \displaystyle \boldsymbol{\int\limits \cos kx \; dx =\frac{1}{k} \cdot \sin kx +C } \\\\\\ \bullet ~ ~ \boldsymbol{\int\limits \sin kx \, dx = -\frac{1}{k} \cdot \cos x + C }[/tex]

[tex]\displaystyle \int\limits \left ( \frac{1}{3}\cos \frac{x}{3} +4 \sin 4x \right ) \, dx = \int \limits \left ( \frac{1}{3} \cos \frac{x}{3} \right ) dx + \int \limit ( 4\sin 4x) dx = \\\\\\ = \frac{1}{3}\cdot \frac{\dfrac{1}{1} }{3} \cdot \sin \frac{x}{3} +4 \cdot \left(-\frac{1}{4} \right ) \cdot \cos 4x =\sin \frac{x}{3} - \cos 4x+C[/tex]

Находим наш  определенный интеграл

[tex]\displaystyle \int\limits^\pi _0 \left ( \frac{1}{3}\cos \frac{x}{3} +4 \sin 4x \right ) \, dx = \left (\sin \frac{x}{3} - \cos 4x \right ) \Bigg |^\pi _0 = \\\\\\ = \sin 60 -\cos 4\pi - (\sin 0 - \cos 0 ) =\frac{\sqrt{3} }{2 } - 1 + 1= \frac{\sqrt{3} }{2}[/tex]