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IfIn=int0^(pi/2)sin^n xdx ,t h e ns howt...

`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`

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`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"):}`
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