求∫(sinx)^6 dx的通解.
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∫ (sinx)^6 dx
=∫ [(sinx)^2]^3 dx
=(1/8)∫ [ 1- cos2x ]^3 dx
= (1/8)∫ [ 1- 3cos2x + 3(cos2x)^2 - (cos2x)^3 ]dx
=(1/8)[ x - (3/2)sin2x] +(3/8)∫ (cos2x)^2 dx-(1/8)∫ (cos2x)^3 dx
=(1/8)[ x - (3/2)sin2x ]+(3/16)∫ (1+cos4x) dx-(1/16)∫ (cos2x)^2 dsin2x
=(1/8)[ x - (3/2)sin2x ] +(3/16)[x+(1/4)sin4x] -(1/16)∫[1- (sin2x)^2] dsin2x
=(1/8)[ x - (3/2)sin2x ] +(3/16)[x+(1/4)sin4x] -(1/16)[sin2x- (1/3)(sin2x)^3] +C
=∫ [(sinx)^2]^3 dx
=(1/8)∫ [ 1- cos2x ]^3 dx
= (1/8)∫ [ 1- 3cos2x + 3(cos2x)^2 - (cos2x)^3 ]dx
=(1/8)[ x - (3/2)sin2x] +(3/8)∫ (cos2x)^2 dx-(1/8)∫ (cos2x)^3 dx
=(1/8)[ x - (3/2)sin2x ]+(3/16)∫ (1+cos4x) dx-(1/16)∫ (cos2x)^2 dsin2x
=(1/8)[ x - (3/2)sin2x ] +(3/16)[x+(1/4)sin4x] -(1/16)∫[1- (sin2x)^2] dsin2x
=(1/8)[ x - (3/2)sin2x ] +(3/16)[x+(1/4)sin4x] -(1/16)[sin2x- (1/3)(sin2x)^3] +C
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