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

[Gwozdzilla]

Homework Statement

Use integration by parts to find a reduction formula for the integral

I_n = ∫^pi/2₀ sinⁿ(6x)dx

when n is a positive integer greater than 1.

Homework Equations

∫udv = uv - ∫vdu

The Attempt at a Solution

Let u = 6x du = 6dx
∫^pi/2₀ sinⁿ(u) (du/6)
(1/6) ∫sin^n-1(u)sinu du
Let v = sin^n-1(u) and dw = sinu du
and dv = (n-1)sin^n-2(u)cosu du and w = -cos(u)
(1/6)(-cos(u)sin^n-1(u) - ∫-cos²(u) (n-1) sin^n-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)∫cos²sin^n-2(u) du
(1/6)(n-1)∫(1-sin²u)sin^n-2u du
(1/6)(n-1)∫sin^n-2u - sinⁿu du
I_n = (1/6)(n-1)(I_n-2 -I_n)
I_n = (1/6)(n-1)(I_n-2) -(1/6)(n-1)(I_n)
I_n + (1/6)(n-1)(I_n)= (1/6)(n-1)(I_n-2)
(1 + (n-1)/6)I_n = (1/6)(n-1)(I_n-2)
((n+5)/6)I_n = (1/6)(n-1)(I_n-2)
I_n = (1/6)(n-1)(I_n-2)(6/(n+5))
I_n = ((n-1)/(n+5))(I_n-2)

This was not one of my answer choices for my homework. Could you please help me see what I'm doing wrong?

 

[Gwozdzilla]

I entered ((n-1)/n)I_n-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.

 

[Ray Vickson]

Homework Statement

Use integration by parts to find a reduction formula for the integral

I_n = ∫^pi/2₀ sinⁿ(6x)dx

when n is a positive integer greater than 1.

Homework Equations

∫udv = uv - ∫vdu

The Attempt at a Solution

Let u = 6x du = 6dx
∫^pi/2₀ sinⁿ(u) (du/6)
(1/6) ∫sin^n-1(u)sinu du
Let v = sin^n-1(u) and dw = sinu du
and dv = (n-1)sin^n-2(u)cosu du and w = -cos(u)
(1/6)(-cos(u)sin^n-1(u) - ∫-cos²(u) (n-1) sin^n-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)∫cos²sin^n-2(u) du
(1/6)(n-1)∫(1-sin²u)sin^n-2u du
(1/6)(n-1)∫sin^n-2u - sinⁿu du
I_n = (1/6)(n-1)(I_n-2 -I_n)
I_n = (1/6)(n-1)(I_n-2) -(1/6)(n-1)(I_n)
I_n + (1/6)(n-1)(I_n)= (1/6)(n-1)(I_n-2)
(1 + (n-1)/6)I_n = (1/6)(n-1)(I_n-2)
((n+5)/6)I_n = (1/6)(n-1)(I_n-2)
I_n = (1/6)(n-1)(I_n-2)(6/(n+5))
I_n = ((n-1)/(n+5))(I_n-2)

This was not one of my answer choices for my homework. Could you please help me see what I'm doing wrong?

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.