Problem: Prove by increasing the inequalities $\dfrac{1-\sin^{2n-1}x}{2n-1}> \dfrac{1-\sin^{2n}x}{2n}>\dfrac{1-\sin^{2n+1}x}{2n+1}$ from $0$ to $\frac{\pi}{2}$ and using the previous example the $p_{n-1}\left(1+\dfrac{2n-1}{2n}p_{n-1}\right)>\dfrac{4n}{\pi}>p_n(p_{n-1})$, where $p_n=\dfrac{3\cdot 5\dots (2n+1)}{2\cdot 4\dots 2n}$.
Is it possible to show the inequality, if we consider $I_n=\int\limits_0^{\pi/2} \sin^n x\, dx$?