Integral Prove the reduction formula?

Question

Integral cos^n(x) dx = (1/n)cos^(n-1)(x)sin(x) + (n-1)/n * Integral cos^(n-2)(x) dx

I did the first integration by parts but now when i'm trying to do it a second time, I don't know how to separate it. Should I let
u= ncos^(n-1)x and then i would get
du=(n-1)/n * cos^(n-2)(x)sin(x)dx
which is starting to look like the formula but it looks like i have an extra sin(x) in there. ????

Answer 1

The trick in this one is not to try to do it a second time but to get a multiple of the original integral as part of the RHS.

Let u = cos^(n-1) x, dv = cos x dx
=> v = sin x, du = (n-1) cos^(n-2)x (-sin x) dx

Let I = ∫ cos^n x dx = ∫ u dv = uv - ∫ v du
= cos^(n-1) x sin x - ∫ -(n-1) cos^(n-2) x sin^2 x
= cos^(n-1) x sin x + (n-1) ∫ cos^(n-2) x (1 - cos^2 x) dx
= cos^(n-1) x sin x + (n-1) [∫ cos^(n-2) x dx - I]
So I + (n-1) I = cos^(n-1) x sin x + (n-1) ∫ cos^(n-2) x dx
=> I = 1/n cos^(n-1) x sin x + ((n-1)/n) ∫ cos^(n-2) x dx

Answer 2

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RE:
Integral Prove the reduction formula?
Integral cos^n(x) dx = (1/n)cos^(n-1)(x)sin(x) + (n-1)/n * Integral cos^(n-2)(x) dx

I did the first integration by parts but now when i'm trying to do it a second time, I don't know how to separate it. Should I let
u= ncos^(n-1)x and then i would get
du=(n-1)/n *...

Answer 3

thank you