Let In=∫0^π/2sin ^n x dx. Then 10 I10 I9 equals

Integration by parts gives the reduction formula $I_n=\dfrac{n-1}{n}I_{n-2}$, so $nI_n=(n-1)I_{n-2}$. Multiply both sides by $I_{n-1}$: $nI_nI_{n-1}=(n-1)I_{n-1}I_{n-2}$. Thus the quantity $P_n=nI_nI_{n-1}$ does not depend on $n$; it is constant. Evaluate at $n=1$: $P_1=1\cdot I_1\cdot I_0=1\cdot1\cdot\dfrac{\pi}{2}=\dfrac\pi2$. Hence $10I_{10}I_9=\dfrac\pi2$. Direct check: $I_{10}=\dfrac{63\pi}{256}$ and $I_9=\dfrac{128}{315}$, whose product times $10$ is $\dfrac{\pi}{2}$. Both factors are positive, so the product is positive, as expected. This recurrence-based product trick is a standard shortcut for Wallis-type integrals and avoids computing either integral separately, which would take many repeated reductions.