If In = ∫-ππ sin(nx)(1+πx)sin(x)dx, n=0,1,2, ..., then what is the value of In?

Let In = ∫-ππ sin(nx)(1+πx)sin(x)dx
Let f(x) = (1+πx)sin(x)
Then In = ∫-ππ sin(nx)f(x)dx
If n=0, I0 = ∫-ππ (1+πx)sin(x)dx = 0
If n=1, I1 = ∫-ππ sin2(x)(1+πx)dx = ∫-ππ ½(1-cos(2x))(1+πx)dx = 10π
If n=2, I2 = ∫-ππ sin(2x)(1+πx)sin(x)dx = 0
If n=3, I3 = ∫-ππ sin(3x)(1+πx)sin(x)dx = 0
In general, In = 0 if n is even, and In = 10π if n is odd.
Therefore, 10∑m=1I2m+1 = 10π