`IfI_(m , n)=int_0^(pi/2)sin^m xcos^n xdx ,` Then show that `I_(m , n)=(m-1)/(m+n)I_(m-2,n)(m ,n in N)` Hence, prove that `I_(m , n)=f(x)={((n-1)(n-3)(m-5)(n-1)(n-3)(n-5))/((m+n)(m+n-2)(m+n-4))pi/4` when both `m` and `n` are even
`((m-1)(m-3)(m-5)(n-1)(n-3)(n-5))/((m+n)(m+n-2)(m+n-4))}`

Answer

NA

Explanation

`I_(m,n)=int_(0)^((pi)/2)sin^(m-1)x(sinx cos^(n)x)dx`
`=[-(sin^(m-1)x cos^(n+1)x)/(n+1)]_(0)^((pi)/2)`
`+int_(0)^((pi)/2)(cos^(n+1))/(n+1)(m-1)sin^(m-2)x cosx dx`
`=((m-1)/(n+1))int_(0)^((pi)/2)sin^(m-2)x cos^(n)x cos^(2)x dx`
`=((m-1)/(n+1))int_(0)^((pi)/2)(sin^(m-2)x cos^(n)x -sin^(m)x cos^(n)x)dx`
`=((m-1)/(n+1))I_(m-2,n)-((m-1)/(n+1))I_(m,n)`
or `(1+(m-1)/(n+1))I_(m,n)=((m-1)/(n+1))I_(m-2,n)`
r `I_(m,n)=((m-1)/(m+n))I_(m-2,n)`
`=((m-1)/(m+n))((m-3)/(m+n-2))((m-5)/(m+n-4))............I_(0,n)` or `I_(1,n)`
According as `m` is even or odd
`I_(0,n)=int_(0)^((pi)/2)cos^(n)x dx` and `I_(1,n)=int_(0)^((pi)/2) sinx cos^(n)x dx=1/(n+1)`
`I_(m,n)={(((m-1)(m-3)(m-5)………(n-1)(n-3)(n-5)………)/((m+n)(m+n-2)(m+n-4)…………….2)(pi)/2,"when both" m "and" n "are even"),(((m-1)(m-3)(m-5)……….(n-1)(n-3)(n-5)..........)/((m+n)(m+n-2)(m+n-4)..........),"otherwise"):}`