Let f(x) be a non-constant twice differentiable function defined on...

Given that, $f(x)=f(1-x)$ On differentiating w.r.t. $x$, we get f^{\prime}(x)=-f^{\prime}(1-x) Let us put $\quad x=\frac{1}{2}$ \Rightarrow \quad 2 f^{\prime}$\left(\frac{1}{2}\right)=$0 \Rightarrow f^{\prime}\left(\frac{1}{2}\right)=$0 Since,$f^{\prime}\left(\frac{1}{2}\right)=0 and $f^{\prime}\left(\frac{1}{4}\right)=0\Rightarrow f^{\prime \prime}(x)=0$ at two points in $[0,1]$. Now, $\int_{-1 / 2}^{1 / 2} f\left(x+\frac{1}{2}\right) \sin x d x=0$ As, $f\left(x+\frac{1}{2}\right) \sin x$ is an odd function which is clear from the following explanation. Let $g(x)=f\left(x+\frac{1}{2}\right) \sin x$, g(-x)=f\left(\frac{1}{2}-x\right) \sin (-x)=-\sin x f\left(1-\left(\frac{1}{2}-x\right)\right)=-\sin x f\left(\frac{1}{2}+x\right)=-g(x) Moreover, $\int_{1/2}^1 f(1-t) e^{\sin (\pi t)} d t=\int_0^{1 / 2} f(u) \cdot e^{\sin \pi u} d u$ where, $1-t=u$.