$$I_n=\int_{\frac n2}^{\frac{n+1}{2}}\dfrac{\sin\left(\pi\sin^2\left(\pi x\right)\right)}{2^{x/2}}\mathrm dx$$
$$I_{n+2}=\int_{\frac n2+1}^{\frac{n+1}{2}+1}\dfrac{\sin\left(\pi\sin^2(\pi x)\right)}{2^{x/2}}\mathrm dx$$
Using $t=x-1$, it becomes :
$$I_{n+2}=\int_{\frac n2}^{\frac n2+1}\dfrac{\sin\left(\pi\sin^2(\pi t+\pi)\right)}{\sqrt2\cdot2^{t/2}}\mathrm dt=\dfrac1{\sqrt2}\int_{\frac n2}^{\frac n2+1}\dfrac{\sin\left(\pi\sin^2(\pi t)\right)}{2^{t/2}}\mathrm dt=\dfrac1{\sqrt2}I_n$$
So, we arrive at : $I_{n+2}=\dfrac1{\sqrt2}I_n
