(sinx)^6(cosX)^4
=(1/32) [(1-cos2x)^3(1+cos2x)^2]
=(1/32) [1-(cos2x)^2]^2(1-cos2x)
=(1/32)[(sin2x)^4-(sin2x)^2(cos2x)]
∫(sin2x)^2 dx = (1/4) ∫(1-cos4x)^2dx = (1/4) [(1-2cos4x+cos4x^2)]dx
=(1/4)[x - (1/2)sin4x + (1/2)x + (1/16)sin8x]
=(3/8)x - (1/8)sin4x + (1/64)sin8x
∫(sin2x)^2(cos2x) dx = ∫(sin 2x)^2 d(sin 2x) = (1/3) (sin 2x)^3
所以，原式：
（1/32)[ (3/8)x - (sin 4x)/8 + (sin8x)/32 - ( sin 2x)^3/3]
代入0和π：得
∫0-pai (sinx)^6*(cosx)^4 dx= 3π／(2^8) = 3π/256 = 0.0368
汗！！！真够麻烦的！
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024