>>43
　∫[0,π] e^x |sin(nx)| dx = Σ[k=0,n-1] (-1)^k ∫[kπ/n, (k+1)π/n] (e^x)sin(x)dx
　　　　= Σ[k=0,n-1] (-1)^k a_k, とおく。

　∫(e^x)sin(x)dx = (e^x){sin(x)-cos(x)}/2 = (e^x)sin(x-π/4)/√2,

a_k = ∫[kπ/n, (k+1)π/n] (e^x)sin(x)dx
　　= [ (e^x){sin(x-π/4)}/√2 ](x=kπ/n, (k+1)π/n)
　　= e^((k+1)π/n){sin((k+1)π/n - π/4)/√2 - e^(kπ/n)sin(kπ/n - π/4)/√2,

>>55
8.5時間ぐらいだが何か