`IfI_n=int_0^(pi/2)sin^n xdx ,t h e ns howt h a tI_n=((n-1)/n)I_(n-2)` Hence, Prove that `I_n=f(x)={((n-1)/n)((n-3)/(n-2))((n-5)/(n-4))ddot(1/2)pi/2ifni se v e n((n-1)/n)((n-3)/(n-2))((n-5)/(n-4))(2/3)1ifni sod d`

Explanation

`I_(n)=int_(0)^((pi)/2 sin^(n)x dx`
`=int_(0)^((pi)/2) sin^(n-1)x sin x dx`
`=[-sin^(n-1)x cosx]_(0)^((pi)/2)+int_(0)^((pi)/2)(n-1)sin^(n-2)x cos^(2)x dx`
`=(n-1)int_(0)^((pi)/2)sin^(n-2)x(1-sin^(2)x)dx`
`=(n-1)int_(0)^((pi)2/2)sin^(n-2)x x-(n-1)int_(0)^((pi)/2)sin^(n)x dx`
or `I_(n)+(n-1)I_(n)=(n-1)I_(n-2)`
or `I_(n)=((n-1)/n)I_(n-2)`
`=((n-1)/n)((n-3)/(n-2))((n-5)/(n-4))...........I_(0)` or `I_(1)`
Accordingly if `n` is even or odd,
`I_(0)=(pi)/2,I_(1)=1`
Hence `I_(n)= {(((n-1)/n)((n-3)/(n-2))((n-5)/(n-4))………(1/2)(pi)/2,"if",n"is even"),(((n-1)/n)((n-3)/(n-2))((n-5)/(n-4))………(2/3)1,"if",n"is odd"):}`