Answer: The reduction formula is \(I_n=n\left(\frac\pi2\right)^{n-1}-n(n-1)I_{n-2}\), for \(n\geqslant2\). The exact value is \(I_4=\frac{\pi^3}{2}-12\pi+24\).

Start with \(I_n=\int_0^{\pi/2}x^n\sin x\,dx\). Integrate by parts with \(u=x^n\) and \(dv=\sin x\,dx\). Then \(du=nx^{n-1}dx\) and \(v=-\cos x\), so

\(I_n=[-x^n\cos x]_0^{\pi/2}+n\int_0^{\pi/2}x^{n-1}\cos x\,dx\).

The boundary term is zero, hence

\(I_n=n\int_0^{\pi/2}x^{n-1}\cos x\,dx\).

Integrate by parts again, now with \(u=x^{n-1}\) and \(dv=\cos x\,dx\). Then

\(I_n=n\left([x^{n-1}\sin x]_0^{\pi/2}-(n-1)\int_0^{\pi/2}x^{n-2}\sin x\,dx\right)\).

Therefore

\(I_n=n\left(\frac\pi2\right)^{n-1}-n(n-1)I_{n-2}\), for \(n\geqslant2\).

Now \(I_0=\int_0^{\pi/2}\sin x\,dx=1\). Hence

\(I_2=2\left(\frac\pi2\right)-2I_0=\pi-2\).

Using the formula with \(n=4\),

\(I_4=4\left(\frac\pi2\right)^3-12I_2=\frac{\pi^3}{2}-12(\pi-2)\).

Thus \(I_4=\frac{\pi^3}{2}-12\pi+24\).