Finding the value of given integral. It was asked to find the correct option(s) for the given integral: $$I_n = \displaystyle\int_{\frac{n}{2}}^{\frac{n+1}{2}}\dfrac{\sin{(\pi(\sin^2{\pi x}}))}{(\sqrt2)^x} \, dx$$
(a)$\dfrac{I_n}{I_{n+4}}=2$
(b)$\dfrac{I_n}{I_{n+4}}=\dfrac{1}{\sqrt2}$
(c)$\dfrac{\displaystyle\sum_{n=0}^\infty I_{8n}}{I_0}=\dfrac{4}{3}$
(d)$ \dfrac{\displaystyle\sum_{n=0}^\infty I_{n}}{I_0}=2$
I tried using King's property to solve this one, but it is not working here, instead it is making it more  complicated.
Any help is appreciated :)

First, notice that for any even integer $2k$
$$I_{n+2k} = \int_{\frac{n}{2}+k}^{\frac{n+1}{2}+k}\frac{\sin\left[\pi \sin^2(\pi x)\right]}{\left(\sqrt{2}\right)^x}dx = \int_{\frac{n}{2}}^{\frac{n+1}{2}}\frac{\sin\left[\pi \sin^2(\pi x + \pi k)\right]}{\left(\sqrt{2}\right)^{x+k}}dx = \frac{1}{\left(\sqrt{2}\right)^k}I_n$$
Selecting $k=2$ gives us
$$\frac{I_n}{I_{n+4}} = \frac{I_n}{\frac{1}{\left(\sqrt{2}\right)^2}I_n} = 2$$
Selecting $k=4n$ gives us
$$\frac{\sum\limits_{n=0}^{\infty}I_{8n}}{I_0} = \frac{\sum\limits_{n=0}^\infty \frac{1}{\left(\sqrt{2}\right)^{4n}}I_0}{I_0} = \sum_{n=0}^\infty \frac{1}{4^n} = \frac{4}{3}$$
Selecting $k=2n$ gives us
$$\frac{\sum\limits_{n=0}^\infty I_n}{I_0} > \frac{\sum\limits_{n=0}^\infty I_{4n}}{I_0} = \frac{\sum\limits_{n=0}^\infty \frac{1}{\left(\sqrt{2}\right)^{2n}}I_0}{I_0} =  \sum_{n=0}^\infty \frac{1}{2^n} = 2$$
because
$$\pi \sin^2(\pi x) \in [0,\pi] \implies I_n > 0$$
Therefore, the only correct choices are options $(a)$ and $(c)$
