Given integral $I=\int_{\frac{3 \pi}{4}}^\pi \sin x-1\left[\frac{4 x}{\pi}\right] \cdot d x$
We known that $\sin x$ lies between -1 to 1 and we can split the values of $\sin x$ between -1 to 0 and 0 to 1 .
If $0 " \sin x^{\prime \prime} 1$ then $[\sin x]=0$ for $x \in[0, \pi]$
Then,
$$
\begin{aligned}
& \mathrm{I}=\int_{\frac{3 \pi}{4}}^\pi[\sin x] d x+\int_{\frac{3 \pi}{4}}^\pi\left[\frac{4 x}{\pi}\right] \cdot d x \\
& \text { If } \frac{3 \pi}{4} \leq x \leq \pi \text { then } \frac{4}{x} \times \frac{3 \pi}{4} \leq \frac{4 \pi}{\pi} \leq \frac{4}{\pi} \times \pi \\
& \Rightarrow \quad 3 \leq \frac{4 x}{\pi} \leq 4
\end{aligned}
$$
So, $\left[\frac{4 x}{\pi}\right]=3$
$$
\begin{aligned}
& I=\int_{\frac{3 \pi}{4}}^\pi(0) d x+3 \int_{\frac{3 \pi}{4}}^\pi d x \\
& I=0+4[x]_{\frac{3 \pi}{4}}^\pi=3\left(\pi-\frac{3 \pi}{4}\right)=\frac{3 \pi}{4}
\end{aligned}
$$