Answer
\[\int_{0}^{2\pi }{\sin x}dx=0\]
Work Step by Step
\[\begin{align}
& \int_{0}^{2\pi }{\sin x}dx \\
& \text{From the graph we can see that the function }\sin x\text{ is odd } \\
& \text{about }x=\pi . \\
& \text{Using the translation, we can express }\int_{0}^{2\pi }{\sin x}dx\text{ as} \\
& \int_{0}^{2\pi }{\sin x}dx=\int_{-\pi }^{\pi }{\sin \left( x+\pi \right)}dx \\
& \text{Using the theorem 5}\text{.4} \\
& \text{If }f\left( x \right)\text{ is odd, }\int_{-a}^{a}{f\left( x \right)dx}=0,\text{ then} \\
& \int_{0}^{2\pi }{\sin x}dx=0 \\
\end{align}\]
