大学高数求不定积分 大学高数不定积分
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答案如图所示
1-(sinx)^3=1-sinx[1-(cosx)^2]=1-sinx+sinx(cosx)^2
设:u=cosx,则du=-sinxdx;又当x=0,π时,u=1,-1
所以:
∫[0,π]:[1-(sinx)^3]dx
=∫[0,π]:[1-sinx+sinx(cosx)^2]dx
=∫[0,π]:dx-∫[0,π]:sinxdx+∫[0,π]:sinx(cosx)^2]dx
=π+∫[1,-1]:du-∫[1,-1]:u^2du
=π-∫[-1,1]:du+∫[-1,1]:u^2du
=π-2+(2/3)
=π-(4/3)
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