>681
求めますた。

∫[0,π] {k･sin(2x)cos(x) - [cos(x)]^2}dx
= [ -(2/3)k･[cos(x)]^3 - (1/2)sin(x)cos(x) -(x/2) ](x=0,π)
= (4/3)k -(π/2),
k = 3π/8,