it does work.
What is your Maclaurin expansion of $\sin(x \sin(t))$ ?
Consider the integral
\begin{equation}
I(x)= \frac{1}{\pi} \int^{\pi}_{0} sin(xsint) dt
\end{equation}
show that
\begin{equation}
I(x)= \frac{2x}{\pi} +O(x^{3})
\end{equation}
as $x\rightarrow0$.
=> I Have used the expansion of McLaurin series of $I(x)$ but did not work.
please help me.
basically as $x \to 0, \sin(x \sin(t))$ can be closely approximated by it's Maclaurin series in x
which in this case is simply $\sin(t) x+O(x^3)$
we drop the $O(x^3)$ part and integrate
$\dfrac 1 \pi \displaystyle{\int_0^{\pi}} \sin(t)x~dt=\dfrac x \pi \cos(t)|_{\pi}^0=\dfrac {2x}{\pi}$