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Set $s=\sin u$, so $ds=\cos udu$, and write $J_n=I_{n,1}(\pi /2,0)$. The integration by substitution gives $J_n={\int }_0^1(1-s^3{)}^{n}ds$. For $n>0$, integration by parts yieldsBoth endpoint terms vanish. Use $s^3=1-(1-s^3)$ inside the remaining integral:ConsequentlyThis is the $p=3$ instance of the power-complement integral recurrence, with initial value $J_0=1$.

Apply the power-complement integral recurrence three times, starting from $I_{0,1}(\pi /2,0)=1$:Each factor corresponds to reducing $n$ by one; the final unreduced integral is just ${\int }_0^{\pi /2}\cos udu=1$.