<s>[INST] Given $U_n=\int_0^\frac{\pi}{2} x\sin^n x dx$, find $\frac{100U_{10}-1}{U_8}$ 
If
$$U_n=\int_0^\frac{\pi}{2} x\sin^n x dx$$
Find $\frac{100U_{10}-1}{U_8}$
Answer: $90$

My Attempt:
I tried applying Integration By Parts, and when that failed, I tried the substitution $x\rightarrow \frac{\pi}{2} -x$ , only to establish a (probably useless) relationship between the $U_n$ and its cosine counterpart:
$$U_n=\frac{\pi}{2}\int_0^{\frac{\pi}{2}} \cos^n \ dx-\int_0^{\frac{\pi}{2}}x\cos ^n x\ dx$$
Any help would be appreciated!
 [/INST] Note
$$U_{n-2} -U_n =\int_0^\frac{\pi}{2} x\sin^{n-2}\cos^2x dx
\overset{IBP}=\frac1{n-1} \int_0^\frac{\pi}{2} x\cos xd(\sin^{n-1}x)\\
= \frac1{n-1}\left( U_n -\frac1n \int_0^\frac{\pi}{2} d(\sin^{n}x)\right)= \frac1{n-1}\left(U_n - \frac1n\right)
$$
Thus, $U_n = \frac{n-1}nU_{n-2} + \frac1{n^2}$ and
$$\frac{100U_{10}-1}{U_8} = \frac{ 100(\frac9{10}U_8+\frac1{100})-1}{U_8}=90$$
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