∫(sinx)^6(cosx)^4dx
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解题思路是逐次降幂，
∫(sinx)^6(cosx)^4dx = (1/32)∫(1-cos2x)^3(1+cos2x)^2dx
= (1/32)∫[1-3cos2x+3(cos2x)^2-(cos2x)^3][1+2cos2x+(cos2x)^2]dx
= (1/32)∫[1-cos2x-2(cos2x)^2+2(cos2x)^3+(cos2x)^4-(cos2x)^5]dx
= (1/32){x-(1/2)sin2x-∫(1+cos4x)dx +∫[1-(sin2x)^2]dsin2x
+(1/4)∫[1+cos4x)^2dx - (1/2)∫[1-(sin2x)^2]^2dsin2x}
= (1/32){x-(1/2)sin2x-x+(1/4)sin4x +sin2x-(1/3)(sin2x)^3
+(1/4)∫[1+2cos4x+(cos4x)^2]dx - (1/2)∫[1-2(sin2x)^2+(sin2x)^4]dsin2x}
= (1/32){x-(1/2)sin2x-x+(1/4)sin4x +sin2x-(1/3)(sin2x)^3
+(1/4)∫[3/2+2cos4x+(1/2)(cos8x)]dx
- (1/2)[sin2x-(2/3)(sin2x)^3+(1/5)(sin2x)^5]}
= (1/32)[3x/8-(1/10)(sin2x)^5+(3/8)sin4x+(1/64)sin8x]
∫(sinx)^6(cosx)^4dx = (1/32)∫(1-cos2x)^3(1+cos2x)^2dx
= (1/32)∫[1-3cos2x+3(cos2x)^2-(cos2x)^3][1+2cos2x+(cos2x)^2]dx
= (1/32)∫[1-cos2x-2(cos2x)^2+2(cos2x)^3+(cos2x)^4-(cos2x)^5]dx
= (1/32){x-(1/2)sin2x-∫(1+cos4x)dx +∫[1-(sin2x)^2]dsin2x
+(1/4)∫[1+cos4x)^2dx - (1/2)∫[1-(sin2x)^2]^2dsin2x}
= (1/32){x-(1/2)sin2x-x+(1/4)sin4x +sin2x-(1/3)(sin2x)^3
+(1/4)∫[1+2cos4x+(cos4x)^2]dx - (1/2)∫[1-2(sin2x)^2+(sin2x)^4]dsin2x}
= (1/32){x-(1/2)sin2x-x+(1/4)sin4x +sin2x-(1/3)(sin2x)^3
+(1/4)∫[3/2+2cos4x+(1/2)(cos8x)]dx
- (1/2)[sin2x-(2/3)(sin2x)^3+(1/5)(sin2x)^5]}
= (1/32)[3x/8-(1/10)(sin2x)^5+(3/8)sin4x+(1/64)sin8x]
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