$\int\limits_0^X[\sin t] d t$ where $x \in(2 n \pi,(4 n+1) \pi, n \in N$ and [.] denotes the greatest integer function is equal to.
Answer & explanation
Correct answer: option 1
$I=\int\limits_0^X[\sin t] d t=\int\limits_0^{2 n \pi}[\sin t] d t+\int\limits_{2 n \pi}^X[\sin t] d t$
$=n \int\limits_0^{2 \pi}[\sin t] d t+\int\limits_{2 n \pi}^X[\sin t] d t$
$=n\left(\int\limits_0^{\pi / 2}[\sin t] d t+\int\limits_{\pi / 2}^{\pi}[\sin t] d t+\int\limits_\pi^{2 \pi}[\sin t] d t\right)+\int\limits_{2 n \pi}^x[\sin t] d t$
$\Rightarrow I=n\left(0+0-\int\limits_\pi^{2 \pi} d t\right)+0=-n \pi$