Решение.

Интегрирование по частям.

[tex]\displaystyle \int\limits_0^{1/3}\ arccos(3x)\, dx=\Big[\ u=arccos(3x)\ ,\ du=-\frac{3\, dx}{\sqrt{1-9x^2}}\ ,\ dv=dx\ ,\\\\\\v=x\ \Big]=x\cdot arccos(3x)\Big|_0^{1/3}-\int \limits_0^{1/3}\ \frac{-3x\, dx}{\sqrt{1-9x^2}}=\frac{1}{3}\cdot arccos1-0+\\\\\\-\frac{1}{6}\int \limits _0^{1/3}\frac{-18\, dx}{\sqrt{1-9x^2}}=\frac{1}{3}\cdot \frac{\pi}{2}-\frac{1}{6}\int \limits _0^{1/3}\frac{d\, (1-9x^2)}{\sqrt{1-9x^2}}=\frac{\pi}{6}-\frac{1}{6}\cdot 2\sqrt{1-9x^2}\ \Big|_0^{1/3}=[/tex]

[tex]\displaystyle =\frac{\pi}{6}-\frac{1}{3}\cdot (\sqrt{0}-\sqrt{1})=\frac{\pi}{6}+\frac{1}{3}=\frac{\pi +2}{6}[/tex]