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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`

Text Solution

Verified by Experts

The correct Answer is:
`pi^(5)`
`I_(m)=int_(0)^(pi)x^(m) sin x dx`
`=[-x^(m)cosx]_(0)^(pi)+n int_(0)^(pi)x^(n-1)cos x dx`
`=pi^(n)+n[x^(n-1)sinx ]_(0)^(pi)-n(n-1)int_(0)^(pi)x^(n-2)sin x dx`
`implies I_(m)=pi^(n)+n.0-n(n-1)I_(n-2)`
Put `n=5`
`I_(5)=pi^(5)-20I_(3)`
`I_(5)+20I_(3)=pi^(5)`
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