If `I_(n)=int_(0)^(pi/2) sin^(x)x dx`, then show that `I_(n)=((n-1)n)I_(n-2)`.
Hence prove that
`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"):}`

Explanation

To solve the problem, we need to show that

\[ I_n = \frac{(n-1)}{n} I_{n-2} \]

and then derive the expressions for \(I_n\) when \(n\) is even and odd.

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