integration by parts

∫sin³xdx
=∫sin²xsinxdx
=∫(1-cos²x)d(cosx)
=cosx-(1/3)cos³x+C, where C is a constant

P.S. This question does not need to use integration by parts

2008-05-31 11:28:51 補充：
First, your answer is wrong, it should be I = 1/3( -sin²xcosx - 2cosx) + C

2008-05-31 11:29:09 補充：
Actually there is some mistakes in my answer, but I will amend it without using Integration by parts, because I insist on the belief that this question can be calculated by other methods with less steps then Integration by parts

2008-05-31 11:29:24 補充：
∫sin³xdx
=∫sin²xsinxdx
=∫(1-cos²x)d(-cosx)
=-∫(1-cos²x)d(cosx)
=-cosx+(1/3)cos³x+C, where C is a constant

2008-05-31 11:29:32 補充：
And my answer is the same as your answer, which can be proved as follows:
-cosx+(1/3)cos³x+C
=-cosx+(1/3)(cos²xcosx)+C
=-cosx+(1/3)[(1-sin²x)cosx]+C
=-cosx+(1/3)cosx-(1/3)sin²xcosx+C
=(-2/3)cosx-(1/3)sin²xcosx+C
=1/3(-sin²xcosx-2cosx)+C

2008-05-31 11:31:42 補充：
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sin3x =sin(x +2x) = sinx cos2x +cosxsin2x =sinx(1-2sin^2x) + cosx(2sinxcosx)
=sinx -2sin^3x + 2sinxcos^2x = sinx -2sin^3x + 2sinx(1-sin^2x)=sinx -2sin^3x +2sinx -2sin^3x = 3sinx - 4sin^3x.
Therefore, sin^3x = (3sinx - sin3x)/4.
Therefore, Ssin^3xdx =1/4 S(3sinx -sin3x)dx = 1/4[-3cosx + cos3x/3] + C.