求高手帮做这几道定积分题
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(1)∫<0,π/2>xsinxdx
=-xcosx|<0,π/2>+∫<0,π/2>cosxdx
=sinx|<0,π/2>
=1.
(2)∫<0,1>x^2ln(1+x)dx
=x^3/3*ln(1+x)|<0,1>-∫<0,1>x^3dx/[3(1+x)]
=(1/3)ln2-(1/3)∫<0,1>[x^2-x+1-1/(x+1)]dx
=(1/3)ln2-(1/3)[x^3/3-x^2/2+x-ln(x+1)]|<0,1>
=(1/3)ln2-(1/3)[1/3-1/2+1-ln2]
=(2/3)ln2-5/18.
(3)∫<0,1>xarctanxdx
=x^2*arctanx|<0,1>-∫<0,1>x^2dx/(1+x^2)
=π/4-∫<0,1>[1-1/(1+x^2)]dx
=π/4-[x-arctanx]|<0,1>
=π/2-1.
=-xcosx|<0,π/2>+∫<0,π/2>cosxdx
=sinx|<0,π/2>
=1.
(2)∫<0,1>x^2ln(1+x)dx
=x^3/3*ln(1+x)|<0,1>-∫<0,1>x^3dx/[3(1+x)]
=(1/3)ln2-(1/3)∫<0,1>[x^2-x+1-1/(x+1)]dx
=(1/3)ln2-(1/3)[x^3/3-x^2/2+x-ln(x+1)]|<0,1>
=(1/3)ln2-(1/3)[1/3-1/2+1-ln2]
=(2/3)ln2-5/18.
(3)∫<0,1>xarctanxdx
=x^2*arctanx|<0,1>-∫<0,1>x^2dx/(1+x^2)
=π/4-∫<0,1>[1-1/(1+x^2)]dx
=π/4-[x-arctanx]|<0,1>
=π/2-1.
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