2/π ∫ π - 0 t^2 cos (nt) dt

Question

how to solve this anyone?

Answer 1

use integration by parts, twice:

u = t^2
du = 2t dt
dv = cos(nt) dt
v = 1/n sin(nt)

∫u dv = uv - ∫v du, so we have
∫t^2 cos(nt) dt = 1/n t^2 sin(nt) - 2/n ∫t sin(nt) dt

Now use u=t, dv=sin(nt)

and do it all again.
Multiply end result by 2/π