Reduction Formulae....

(i) 'm' is a +ve integer and n ≥ 2

Proof:

Im, n = ∫ sinm cos n x dx
= ∫ (sinm x cosn – 1 x) cos x dx
= sinm x cosn – 1 x sin x – ∫ [sinm x (n – 1) cosn – 2 x (– sin x) + cosn – 1 x m sinm – 1 x cos x] sin x dx
= sinm + 1 x cosn – 1 x + (n – 1) ∫ sinm x cosn – 2 x sin2 x dx – m ∫ sinm x cosn x dx
= sinm + 1 x cosn – 1 x + (n – 1) ∫ sinm x cosn – 2 x(1 – cos2 x)dx – m Im, n
= sinm + 1 x cosn – 1 x + (n – 1) ∫ sinm x cosn – 2 x dx – (n – 1) ∫ sinm x cosn x dx – m Im, n
= sinm + 1 x cosn – 1 x + (n – 1) Im, n – 2 – (n – 1)Im, n – m Im, n
(m + n) Im, n = sinm + 1 x cosn – 1 x + (n – 1) Im, n – 2
⇒ Im, n =

(ii) Im, n =   m ≥ 2 and 'n' is a +ve integer

Note:   n ≥ 2, m ≥ 2