Answer
$\dfrac{1}{2}$
Work Step by Step
When the function $f(x)$ has minimum and maximum values defined on $[p,q]$ then the maximum-minimum inequality tells that
$min f(q-p) \leq \int_p^q f(x) dx \leq max f(q-p)$
As we are given that $f(x)=\sin x \leq x$ for $x \geq 0$
$\implies \sin x-x \leq 0$ for $x \geq 0$
or, $ \int_0^1 (\sin x-x) dx \leq 0$
This implies that $\int_0^1 \sin x dx \leq \int_0^1 x dx$
Thus, $\int_0^1 \sin x dx \leq \dfrac{1}{2}$
So, we get the upper bound: $\dfrac{1}{2}$