Question

Give reduction formula for ∫ sinⁿ x dx.

Answer 1

Let I_n = ∫ sinⁿ x dx 

 = ∫ sin^{n – 1} x .sin x dx 

 = sin^n–1 x ∫ sin x dx – ∫ (n – 1)sin^n–2 x (– cos² x)dx 

 = sin^{n – 1} x(– cos x) + (n – 1) ∫ sin^{n– 2} x (1 – sin² x)dx 

 = sin^{n – 1} x(– cos x) + (n – 1)I_n–2 – I_n 

 Thus, (1 + n – 1)I_n = – cos x .sin^n–1 x + (n – 1)I_{n – 2} 

nI_n = – cos x .sin^n–1 x + (n – 1)I_{n – 2} 

 I_n = ((cos x .sin^{n - 1}x)/n) + ((n – 1)/n)I_{n – 2} 

which is the required reduction formula.

Answer 2

Let I_n​ = ∫ sinⁿx dx

⇒ I_n​ = ∫ sin x ⋅ sin^n-1x dx

Integrating by parts, we have

I_n ​= −cos x(sinⁿ⁻¹x) - ∫ (-cos x)(n−1)sin^n-2 x cos x dx

I_n​ = −sin^n-1x cos x + (n-1) ∫ cos² x sin^n-2 x dx

I_n​ = −sin^n-1x cos x + (n-1) ∫ sinⁿ⁻²x (1−sin²x)dx

I_n​ = −sin^n-1x cos x +(n-1) ∫ sin^n-2x dx - (n−1) ∫ sinⁿ x dx

I_n​ = −sin^n-1x cos x +(n−1) ∫ sin^n-2x dx - (n−1)I_n​

I_n​ + (n−1)I_n​ = −sin^n-1x cos x +(n−1) ∫ sin^n-2 x dx

\(I_n = -\frac{1}{n}sin^{n-1}x\,cos\,x+ \frac{(n-1)}{n} \int \,sin^ {n-2}x\,dx\)

Hence,

\(\int\,sin^nx\,dx = -\frac{1}{n}sin^{n-1}x\,cos\,x+ \frac{(n-1)}{n} \int \,sin^ {n-2} x\,dx.\)