= Solution
Set $s=\sin u$, so $ds=\cos u\,du$, 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> yields
$$
J_n=[s(1-s^3)^n]_0^1+3n\int_0^1s^3(1-s^3)^{n-1}\,ds.
$$
Both endpoint terms vanish. Use $s^3=1-(1-s^3)$ inside the remaining <integral>:
$$
J_n=3n(J_{n-1}-J_n).
$$
Consequently
$$
\boxed{I_{n,1}(\pi/2,0)=\frac{3n}{3n+1}I_{n-1,1}(\pi/2,0),\qquad
f(n)=\frac{3n}{3n+1}}.
$$
This is the $p=3$ instance of the <power-complement integral recurrence>, with initial value $J_0=1$.
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