The equation $\sin x·x\cos x=0$ has at least one root in the interval.

Let $f\left(x\right)=\int (\sin x+x\cos x)dx$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\int \mathrm{sinx} \mathrm{dx} + \int \mathrm{xcosx} \mathrm{dx}$

By applying integration by parts we get,

$$\begin{gathered} \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=-\mathrm{cosx}+\mathrm{x}\int \mathrm{cosx} \mathrm{dx}-\int \frac{d\mathrm{x}}{d\mathrm{x}}\int \mathrm{cosx} \mathrm{dx} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{xsinx} \end{gathered}$$

As we observe that $f\left(0\right)=f\left(\pi \right)=0$

Since $0$ and $\pi$ are two roots of $f\left(x\right)=0$.

- Consequently, $f^{\text{'}}\left(x\right)=0$, i.e. $\sin x·x\cos x=0$ has at least one root in $(0,\pi )$ by applying the Rolle's theorem.

Hence, option-$2$ is the correct answer.