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ä¸å®ç§¯åâ«sin²xdx
解ï¼åå¼=â«[(1-cos2x)/2]dx=(1/2)x-(1/2)â«cos2xdx=(1/2)x-(1/4)â«cos2xd(2x)=(1/2)x-(1/4)sin2x+C
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³äºâ«sinⁿxdxæ
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â«sinⁿxdx=-(sinⁿֿ¹xcosx)/n+[(n-1)/n]â«sinⁿֿ²xdx.
â«sin⁴xdx=-â«sin³xd(cosx)=-[sin³xcosx-3â«cos²xsin²xdx]=-sin³xcosx+3[â«(1-sin²x)sin²xdx]
=-sin³xcosx+3â«sin²xdx-3â«sin⁴xdx
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移项æ4â«sin⁴xdx=-sin³xcosx+3â«sin²xdx=-sin³xcosx+3[(1/2)x-(1/4)sin2x]+C
=-sin³xcosx-(3/4)sin2x+(3/2)x+C=-sin³xcosx-(3/2)sinxcosx+(3/2)x+C
=-sinxcosx(sin²x+3/2)+(3/2)x+C₁
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â«sin⁴xdx=-(1/4)sinxcosx(sin²x+3/2)+6x+4C₁=-(1/8)sin2x(sin²x+3/2)+6x+C.