问一道高等数学积分题∫(x^3*arccosx/√(1-x^2))dx

∫(x^3*arccosx/√(1-x^2))dx
令t=arccosx t∈(0,π)
则原式=∫(cost)^3*t/sintdcost=-∫(cost)^3*tdt
=-∫(cost)^2cost*tdt=-∫[1-(sint)^2]cost*tdt
=-∫[1-(sint)^2]cost*tdt
=-∫cost*tdt+∫(sint)^2cost*tdt+∫sintdt-∫sintdt
=-∫(sint+cost*t)dt+∫(sint)^2cost*tdt+∫sintdt
=-tsint-cost+∫(sint)^2cost*tdt
=-tsint-cost+∫sin2tsint*t/2dt
=-tsint-cost-∫sin2t(-sint)*t/2dt+∫sin2tcost/2dt
-∫sin2tcost/2dt+∫cos2tcost*t-∫cos2tcost*t
=-tsint-cost-(∫sin2t(-sint)*t/2dt+∫sin2tcost/2dt
+∫cos2tcost*tdt)+∫sin2tcost/2dt+∫cos2tcost*tdt
=-tsint-cost-sin2tcost*t/2+∫sin2tcost/2dt+∫cos2tcost*tdt
=-tsint-cost-sin2tcost*t/2+∫sintcostcostdt
+∫(cost^2-sint^2)cost*tdt
=-tsint-cost-sin2tcost*t/2-∫costcostdcost
+∫(2cost^2-1)cost*tdt
=-tsint-cost-sin2tcost*t/2-cost^3/3+2∫(cost)^3*tdt-∫cost*tdt
=-tsint-cost-sin2tcost*t/2-cost^3/3+2∫(cost)^3*tdt-∫cost*tdt
-∫sintdt+∫sintdt
=-tsint-cost-sin2tcost*t/2-cost^3/3+2∫(cost)^3*tdt
-∫(cost*t+sint)dt+∫sintdt
=-tsint-cost-sin2tcost*t/2-cost^3/3+2∫(cost)^3*tdt
-tsint-cost
这时得
-3∫(cost)^3*tdt=-2tsint-2cost-cost^3/3-sin2tcost*t/2
则-∫(cost)^3*tdt=（-2tsint-2cost-sin2tcost*t/2-cost^3/3）/3
在把t代换成x即可
在网吧里 没有纸笔 一步一步想出来的 自己都觉得烦似乎走了弯路不知道中间有没有错 我现在只能写到这啦…………………………