# Reduction Formula for Integral (sin(6x))^n

1. Sep 3, 2014

### Gwozdzilla

1. The problem statement, all variables and given/known data
Use integration by parts to find a reduction formula for the integral

In = ∫pi/20 sinn(6x)dx

when n is a positive integer greater than 1.

2. Relevant equations

∫udv = uv - ∫vdu

3. The attempt at a solution

Let u = 6x du = 6dx
pi/20 sinn(u) (du/6)
(1/6) ∫sinn-1(u)sinu du
Let v = sinn-1(u) and dw = sinu du
and dv = (n-1)sinn-2(u)cosu du and w = -cos(u)
(1/6)(-cos(u)sinn-1(u) - ∫-cos2(u) (n-1) sinn-2(u) du
When the left hand side of the equation is evaluated from 0 to pi/2, it is found to equal 0.
(1/6)(n-1)∫cos2sinn-2(u) du
(1/6)(n-1)∫(1-sin2u)sinn-2u du
(1/6)(n-1)∫sinn-2u - sinnu du
In = (1/6)(n-1)(In-2 -In)
In = (1/6)(n-1)(In-2) -(1/6)(n-1)(In)
In + (1/6)(n-1)(In)= (1/6)(n-1)(In-2)
(1 + (n-1)/6)In = (1/6)(n-1)(In-2)
((n+5)/6)In = (1/6)(n-1)(In-2)
In = (1/6)(n-1)(In-2)(6/(n+5))
In = ((n-1)/(n+5))(In-2)

2. Sep 3, 2014

### Gwozdzilla

I entered ((n-1)/n)In-2 as my answer and got it correct. My guess as to why this is right is because u is actually a function itself, so when I take the derivative of v, it needs to also be multiplied by du (from my original u-substitution), which is 6. When my original answer is multiplied by 6, I get the correct answer.

3. Sep 3, 2014

### Ray Vickson

Be careful:
$$I_n = \int_0^{\pi/2} \sin^n(6x) \, dx = \frac{1}{6}\int_0^{3 \pi} \sin^n(u) \, du$$
You had the u-integral going from u = 0 to u = π/2, which is incorrect.