Answer
\begin{aligned}
\int_{0}^{\pi} 3 \sin \frac{x}{3} d x
=\frac{9}{2}
\end{aligned}
Work Step by Step
Given $$\int_{0}^{\pi} 3 \sin \frac{x}{3} d x $$
So, we have
\begin{aligned}
I&=\int_{0}^{\pi} 3 \sin \frac{x}{3} d x\\
&=9 \int_{0}^{\pi} \sin \frac{x}{3} \cdot \frac{1}{3} d x\\
&=9\left[-\cos \frac{x}{3}\right]_{0}^{\pi}\\
&=9\left(-\cos \frac{\pi}{3}+\cos 0\right)\\
&=9\left(-\frac{1}{2}+1\right)\\
&=\frac{9}{2}
\end{aligned}
