Question

Give reduction formula for ∫ sin^mx sin nx dx.

Answer 1

Let I_m,n = ∫ sin^m x sin nx dx 

= sin^mx ∫ sin nx dx 

– ∫ (msin^m–1 x .cos x .– (cos nx)/n) dx 

= – (sin^mx .cos nx)/n

+ (m/n) ∫ sin^m–1 x cos x cos nx dx 

= – (sin^mx . cos nx)/n

+ (m/n) ∫ sin^{m– 1} x (cos(n – 1))x – sin nx sin x)dx

= – (sin^mx . cos nx)/n

+ (m/n) ∫ sin^m–1 x (cos(n – 1)) x dx 

– (m/n) ∫ sin^mx sin nx dx 

(1 + (m/n))I_{m, n} = – (sin^mx . cos nx)/n

+ (m/n)I_{m – 1,n –1}

I_{m, n} = – (sin^mx . cos nx)/(m + n) + m/(m + n)I_{m –1, n – 1} 

Comments

how  ∫ sinm– 1 x (cos(n – 1))x becomes Im-1,n-1 when given is I=sin^mxsinnx?