Solution
= Solution
Subtract the integrands first, using $\cos u-\cos^3u=\cos u\sin^2u$. Then set $s=\sin u$:
$$
I_{n,1}(\pi/2,0)-I_{n,3}(\pi/2,0)
=\int_0^1s^2(1-s^3)^n\,ds.
$$
The further <integration by substitution> $w=s^3$ makes $dw=3s^2ds$, so
$$
\boxed{I_{n,1}(\pi/2,0)-I_{n,3}(\pi/2,0)
=\frac13\int_0^1(1-w)^n\,dw=\frac1{3(n+1)}}.
$$
This also holds at $n=0$.