If $x \in\left[(4 n+1) \frac{\pi}{2},(4 n+3) \frac{\pi}{2}\right]$ and $n \in N$, then the value of $\int\limits_0^x[\cos t] d t$, is

$(2 n+1) \frac{\pi}{2}-x$

The graph of $f(x)=[\cos x]$ is as shown in Figure.

Clearly, it is a periodic function with period $2 \pi$.

Now,

$\int\limits_0^x[\cos t] d t=\int\limits_0^{(4 n+1)}[\cos t] d t+\int\limits_{(4 n+1) \pi / 2}^x[\cos t] d t$

$\Rightarrow \int\limits_0^x[\cos t] d t=\int\limits_0^{2 n \pi+\pi / 2}[\cos t] x d t+\int\limits_{2 n \pi+\pi / 2}^x[\cos t] d t$

$\Rightarrow \int\limits_0^x[\cos t] d t=\int\limits_0^{2 n \pi}[\cos t] d t+\int\limits_{2 n \pi}^{2 n \pi+\pi / 2}[\cos t] d t$

$\Rightarrow \int\limits_0^x[\cos t] d t=\int\limits_0^{2 n \pi}[\cos t] d t+\int\limits_0^{\pi / 2}[\cos t] d t+\int\limits_{2 n \pi+\pi / 2}^x[\cos t] d t$

$\Rightarrow \int\limits_0^x[\cos t] d t=n \int\limits_0^{2 \pi}[\cos t] d t+\int\limits_0^{\pi / 2}[\cos t] d t + \int\limits_{2 n \pi+\pi / 2}^x[\cos t] d t$

$\Rightarrow \int\limits_0^x[\cos t] d t=n\left[\int\limits_0^{\pi / 2} 0 d t+\int\limits_{\pi / 2}^{3 \pi / 2}(-1) d t+\int\limits_{3 \pi / 2}^{2 \pi} 0 d t\right] +\int\limits_0^{\pi / 2} 0 d t+\int\limits_{2 n \pi+\pi / 2}^x-1 d t$

$\Rightarrow \int\limits_0^x[\cos t] d t=n\left[-\left(\frac{3 \pi}{2}-\frac{\pi}{2}\right)\right]-\left[x-\left(2 n \pi+\frac{\pi}{2}\right)\right]$

$\Rightarrow \int\limits_0^x[\cos t] d t=n \pi+\frac{\pi}{2}-x=(2 n+1) \frac{\pi}{2}-x$