Hello Krazy G,
We are given:
$$I_n=\int_{0}^{\frac{\pi}{2}} x^n\sin(x)\,dx$$ where $$2\le n$$
and we are asked to derive the given reduction of order formula:
$$I_n=n\left(\frac{\pi}{2} \right)^{n-1}-n(n-1)I_{n-2}$$
To begin, let's use integration by parts where:
$$u=x^n\,\therefore\,du=nx^{n-1}\,dx$$
$$dv=\sin(x)\,dx\,\therefore\,v=-\cos(x)$$
Hence, we may state:
$$I_n=\left.-x^n\cos(x) \right|_{0}^{\frac{\pi}{2}}+n\int_{0}^{\frac{\pi}{2}} x^{n-1}\cos(x)\,dx$$
$$I_n=-\left(\frac{\pi}{2} \right)^n\cos\left(\frac{\pi}{2} \right)+0^n\cos(0)+n\int_{0}^{\frac{\pi}{2}} x^{n-1}\cos(x)\,dx$$
$$I_n=n\int_{0}^{\frac{\pi}{2}} x^{n-1}\cos(x)\,dx$$
Let's use integration by parts again where:
$$u=nx^{n-1}\,\therefore\,du=n(n-1)x^{n-2}$$
$$dv=\cos(x)\,dx\,\therefore\,v=\sin(x)$$
Hence, we may state:
$$I_n=\left.nx^{n-1}\sin(x) \right|_{0}^{\frac{\pi}{2}}-n(n-1)\int_{0}^{\frac{\pi}{2}}x^{n-2}\sin(x)\,dx$$
Observing that $$\int_{0}^{\frac{\pi}{2}}x^{n-2}\sin(x)\,dx=I_{n-2}$$ we may write:
$$I_n=n\left(\frac{\pi}{2} \right)^{n-1}\sin\left(\frac{\pi}{2} \right)-n\cdot0^{n-1}\sin(0)-n(n-1)I_{n-2}$$
$$I_n=n\left(\frac{\pi}{2} \right)^{n-1}-n(n-1)I_{n-2}$$
Shown as desired.