第1个回答 2018-06-23
反复利用降幂公式:
(cosx)^8 = [(cosx)^2]^4 = (1/16)(1+cos2x)^4
= (1/16)[1 + 4cos2x + 6(cos2x)^2 + 4(cos2x)^3 + (cos2x)^4]
= (1/16)[1 + 4cos2x + 3 + 3cos4x + 4(cos2x)^3 + (1/4)(1+cos4x)^2]
= (1/16)[17/4 + 4cos2x + (7/2)cos4x + 4(cos2x)^3 + (1/4)(cos4x)^2]
= (1/16)[35/8 + 4cos2x + (7/2)cos4x + 4(cos2x)^3 + (1/8)cos8x]
I = ∫<0, π/2>(cosx)^8dx
= (1/16)∫<0, π/2>[35/8 + 4cos2x + (7/2)cos4x + 4(cos2x)^3 + (1/8)cos8x]dx
= (1/16)∫<0, π/2>[35/8 + 4cos2x + (7/2)cos4x + (1/8)cos8x]dx
+ (1/16)∫<0, π/2> 4(cos2x)^3 dx
= (1/16)[35/8 + 2sin2x + (7/8)sin4x + (1/64)sin8x]<0, π/2>
+ (1/8)∫<0, π/2>[1- (sin2x)^2]dsin2x
= 35π/256 +(1/8)[sin2x-(1/3)(sin2x)^3]<0, π/2> = 35π/256.
也可直接代瓦利斯公式得
I<8> = ∫<0, π/2>(cosx)^8dx = (7/8)(5/6)(3/4)(1/2)(π/2) = 35π/256