Let $\mathrm{f}\left(x\right)$ be a non-constant twice differentiable function defined on $(-\infty ,\infty )$ such that $\mathrm{f}\left(x\right)=f(1-x)$and ${\mathrm{f}}^1\left(\frac{1}{4}\right)=0.$Then,

Given that, $f\left(x\right)=f(1-x)$

 On differentiating w.r.t. $\mathrm{x},$we get $f^1\left(x\right)=-f^1(1-x)$

            Let us put $x=\frac{1}{2}$

            $\Rightarrow 2f^1\left(\frac{1}{2}\right)=0\Rightarrow f^1\left(\frac{1}{2}\right)=0$

            Since, $f^1\left(\frac{1}{2}\right)=0$ and $f^1\left(\frac{1}{4}\right)=0$

            $\Rightarrow f^{"}\left(x\right)=0$at two points in $[0,1]$.

            Now, $\underset{-1/2}{\overset{1/2}{\int }}f(x+\frac{1}{2})\sin xdx=0$

            As $f(x+\frac{1}{2})\sin x$is an odd function which is clear from the following explanation.

            The following explanation.

            Let $g\left(x\right)=f(x+\frac{1}{2})\sin x,$

            $\Rightarrow g(-x)=f(\frac{1}{2}-x)\sin (-x)=-\sin xf(1-(\frac{1}{2}-x))$

            $=-\sin xf(\frac{1}{2}+x)=-g\left(x\right)$

            Moreover, $\underset{1/2}{\overset{1}{\int }}f(1-t)e^{\sin \left(\pi t\right)}dt=\underset{0}{\overset{1/2}{\int }}f\left(u\right).e^{\sin \pi u}du$

            Where $1-t=u$