Determine a positive integer `n` such that `int_0^(pi/2)x^nsinx dx=3/4(pi^2-8)`

Explanation

Let `I_(n)=int_(0)^(pi//2) x^(n)sin x dx`
Integrate by parts and choose `sinx` as the second function.
Therefore `I_(n)=[x^(n)(-cosx)]_(0)^(pi//2)-int_(0)^(pi//2)nx^(n-1)(-cosx)dx`
`=0+n int_(0)^(pi//2)x^(n-1)cosxdx`
Again integrating by parts, we get
`I_(n)=n{x^(n-1)sinx}_(0)^(pi//2)-n(n-1)int_(0)^(pi//2)x^(n-2)sinxdx`
`=n((pi)/2)^(n-1)-n(n-1)I_(n-2)`
R.H.S contains `pi^(2)`. Therefore put `n=3`
`:.I_(3)=3((pi)/2)^(2)-3xx2I_(1)`
`=(3pi^(2))/4-6int_(0)^(pi/2)xsinxdx`
`=(3pi^(2))/4-6{x(-cosx)+sinx}_(0)^(pi//2)`
`=(3pi^(2))/4-6{1}`
`=3/4(pi^(2)-8)` which is true
Hence `n=3`.