(4)=∫(-π/6到π/6)usinu/cosudsinu
=∫usinudu
=-∫udcosu
=-ucosu+∫cosudu
=sinu-ucosu
=1-√3π/6
(5)函数周期性质
=∫(-π到π)sin²2x+2sin2xcosx+cos²xdx
定积分偶倍奇零
=2∫(0到π)sin²2x+cos²xdx
重复以上
=4∫(0到π/2)sin²2x+cos²xdx
=2∫(1-cos4x+cos2x+1)dx
=4x-sin4x/2+sin2x
=2π
(6)=∫e^x/(e^2x+e)dx
=∫e^(1/2)e^(x-1/2)/e(e^(2x-1)+1)dx
=(1/√e)arctan(e^(x-1/2))
=(arctan(√e)-arctan(1/√e))/√e
(7)=∫(0到ln2)u/e^2ud(e^u-1)
=∫ue^(-u)du
=-ue^(-u)-e^(-u)
=-ln2/2-1/2-(-1)
=(1-ln2)/2追问
=∫usinudu
=-∫udcosu
=-ucosu+∫cosudu
=sinu-ucosu
=1-√3π/6
(5)函数周期性质
=∫(-π到π)sin²2x+2sin2xcosx+cos²xdx
定积分偶倍奇零
=2∫(0到π)sin²2x+cos²xdx
重复以上
=4∫(0到π/2)sin²2x+cos²xdx
=2∫(1-cos4x+cos2x+1)dx
=4x-sin4x/2+sin2x
=2π
(6)=∫e^x/(e^2x+e)dx
=∫e^(1/2)e^(x-1/2)/e(e^(2x-1)+1)dx
=(1/√e)arctan(e^(x-1/2))
=(arctan(√e)-arctan(1/√e))/√e
(7)=∫(0到ln2)u/e^2ud(e^u-1)
=∫ue^(-u)du
=-ue^(-u)-e^(-u)
=-ln2/2-1/2-(-1)
=(1-ln2)/2追问
第七到错了
可不可以把，每个题的过程在清楚一点啊，看太不懂
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