Q.405. Reduction Formula in Integration.

Question 405.
If I(n) = ∫ (from 0 to pi/2) x sin^n x dx,
prove that I(n) = [ (n-1)/n ] I(n-2) + 1/n^2.

Answer 405.
I(n)
= ∫ (from 0 to π/2) x sin^n x dx
= ∫ (from 0 to pi/2) x sin^(n-1) x sinx dx
= [x sin^(n-1)x (-cosx)] ... (0 to π/2)
    + ∫ (from 0 to π/2) [sin^(n-1) x
    + x (n-1) sin^(n-2) x cosx] cosx dx
= ∫ (from 0 to π/2) [sin^(n-1) x cosx dx
    + (n-1) ∫ (from 0 to π/2) x sin^(n-2) * (1 - sin^2 x) dx
= (1/n) sin^n x ... (x = 0 to π/2) + (n-1) [I(n-2) - In]
=> (1 + n - 1) In = 1/n + (n-1) I(n-2)
=> In = 1/n^2 + [(n-1)/n] I(n-2).

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