reduction formula?

u = x^n

dv = cosx * dx

du = nx^(n-1)

v = sinx

I_n = ∫ x^n cos(x) dx

= x^n * sinx - n* ∫x^(n-1) * sinx * dx

= 0 - n*∫x^(n-1) * sinx * dx after applying limits

= -n * [-(x)^(n-1)*cosx + (n-1)*∫ x^(n-2)* cosx * dx]

= -n * [π^(n-1) + (n-1)*I_n-2]

= -nπ^(n-1) - n*(n-1)*I_n-2

So starting with n = 5

I_5 = -5*π^4 - 20*I_3

I_3 = -3*π^2 - 6*I_1

I_1 = - n*∫x^(n-1) * sinx * dx from an earlier line

= ∫ -sinx * dx

= cosx = -2 after limits

So I_3 = -3*π^2 + 12

I_5 = -5*π^4 + 60*π^2 - 240

= -134.869

If the limits are π first then 0 second,

I_5 = +134.869

Please check over my work in case I made any errors. It's easy to make mistakes in questions like this.

It's very hard to do this by repeatedly applying the formulas. It's pretty easy using a table method of integration by parts. You just have to continually differentiate x^5 and continually integrate cosx. Then you multiply across, alternating plus and minus. All the sin terms can be ignored since sin 0 = sin pi = 0. You end up with

(5x^4-6x^2+120)*cosx from 0 to pi.

This becomes -5 pi^4 + 60 pi^2 -240, which is what the other answerer got.