Q.403. Reduction formula in integration

Question 403.
If I(n) = ∫ (from 0 to pi/2) cos^n x sin nx dx,
prove that 2I(n+1) = I(n) + 1/(n+1).

Answer 403.
I(n+1)
= ∫ (x=0 to π/2) cos^(n+1)x sin(n+1) x dx
Using integration by parts
= [cos^(n+1)x ∫ sin(n+1)x dx] ... (x=0 to π/2)
          - ∫ (x=0 to π/2) [d/dx (cos^(n+1)x) ∫ sin(n+1)x dx] dx
= - 1/(n+1) [cos^(n+1)x cos(n+1)x] ... (x=0 to π/2)
   - ∫ (x=0 to π/2) [1/(n+1)cos^n x * (-sinx) * 1/(n+1) * (-cos(n+1)x) dx
= 1/(n+1) - ∫ (x=0 to π/2) cos^n x cos(n+1)x sinx dx ... ( 1 )

sin(nx) = sin[(n+1)x - x] = sin(n+1)x cosx - cos(n+1)x sinx
=> cos(n+1)x sinx = sin(n+1)x cosx - sin(nx)

Plugging this result in ( 1 ),
I(n+1)
= 1/(n+1) - ∫ (x=0 to π/2) cos^n x *[sin(n+1)x cosx - sin(nx)] dx
= 1/(n+1) - ∫ (x=0 to π/2) cos^(n+1)x sin(n+1) x dx + ∫ (x=0 to π/2) cos^n x sin(nx) dx
= 1/(n+1) - I(n+1) + I(n)
=>
2I(n+1) = I(n) + 1/(n+1).

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