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!
 A: 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
$$\frac{100U_{10}-1}{U_8} = \frac{ 100(\frac9{10}U_8+\frac1{100})-1}{U_8}=90$$
A: You can use this I will edit this soon
Now you can plug values of n into last part then you can get your answer.
A: Given,$$u_{n}=\int^{\frac{\pi}{2}}_{0} x \sin^{n}x\ dx=\int^{\frac{\pi}{2}}_{0} (x \cdot \sin x) \sin^{n-1}x\ dx$$
Using Integration by parts:
$$u_{n}=\left[\sin^{n-1}x(-x \cdot \cos x + \sin x)\right]^{\frac{\pi}{2}}_{0}-\int^{\frac{\pi}{2}}_{0} (-x \cdot \cos x + \sin x) (n-1)\sin^{n-2}x \cos x\ dx$$
$$\Rightarrow u_{n}=1+(n-1)\int^{\frac{\pi}{2}}_{0} (x \sin^{n-2}x\cdot \cos^{2}x - \sin^{n-1}x \cos x)\ dx$$
$$\Rightarrow u_{n}=1+(n-1)\int^{\frac{\pi}{2}}_{0} (x \sin^{n-2}x\cdot (1-\sin^{2}x)\ dx - (n-1)\int^{\frac{\pi}{2}}_{0}\sin^{n-1}x \cos x)\ dx$$
$$\Rightarrow u_{n}=1+(n-1)\int^{\frac{\pi}{2}}_{0} x \sin^{n-2}x\ dx- (n-1)\int^{\frac{\pi}{2}}_{0} x\sin^{n}x)\ dx - (n-1) \cdot \frac{1}{n}$$
$$\Rightarrow u_{n}=1+(n-1)u_{n-2}- (n-1)u_{n} - 1+ \frac{1}{n}$$
$$\Rightarrow n \cdot u_{n}=(n-1)u_{n-2}+ \frac{1}{n}$$
$$\Rightarrow \bbox[5px,border:2px solid red]
{u_{n}=\frac{n-1}{n}u_{n-2}+ \frac{1}{n^{2}}}$$
I think you can proceed from here.
