Answer
$$\textbf{True}$$
Work Step by Step
$$\int_a^{b+2\pi}sin(x)dx = \int_a^{b}sin(x)dx + \int_b^{b+2\pi}sin(x)dx $$
Now, $$\int_b^{b+2\pi}sin(x)dx =[ -cos(b+2\pi)+cos(b)]=-cos(b)+cos(b)=0$$
So,
$$\int_a^{b+2\pi}sin(x)dx = \int_a^{b}sin(x)dx $$
