Q.400. Reduction formula in integration

Question 400.
Obtain a reduction formula for :
I(n) = ∫ (from 0 to pi/2) x^n sin x dx .....where n > 1
and hence evaluate I(3).
=> I(n) + n(n-1) I(n-2) = n (π/2)^(n-1)

Answer 400.
Integrating by parts,
I(n)
= [-x^n cosx] (x=0 to π/2) + n ∫x^(n-1) cosx dx ... (x=0 to π/2)
= 0 + n [ {x^(n-1) sinx} (x=0 to π/2) - (n-1) ∫ x^(n-2) sinx dx ] ... (x=0 to π/2)
= n (π/2)^(n-1) - n(n-1) I(n-2)

I(3)
= 3(π/2)^2 - 6 I(1)
= 3(π/2)^2 - 6 ∫ (from 0 to π/2) x sin x dx ... ( 1 )

∫ (from 0 to π/2) x sin x dx
= - x cosx + ∫ cosx dx ... (x=0 to π/2)
= - x cosx + sinx ... (x=0 to π/2)
= 0 + 1

Plugging in ( 1 ),
I(3) = 3(π/2)^2 - 6.

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