Question

求积分。谢谢

Answer 1

∫x^2arcsinx/√(1-x^2)dx
=∫x^2arcsinxd(arcsinx)
令t=arcsinx
x=sint
x^2=sin^2t=1/2(1-cos2t)
上式=∫1/2(1-cos2t)tdt
=∫1/2tdt-∫1/2cos2t*tdt
=1/4t^2+C-1/2∫tcos2tdt
u=t du=dt
dv=cos2tdt
v=1/2sin2t
∫tcos2tdt=t*1/2sin2t-∫1/2sin2tdt
=t/2sin2t+1/4cos2t
1/4t^2+C-1/2∫tcos2tdt
=1/4t^2-1/2(t/2sin2t+1/4cos2t)+C
=1/4t^2-t/4sin2t-1/8cot2t+C
=1/4(arcsinx)^2-1/4arcsinxsin2(arcsinx)-1/8cos2(arcsinx)+C
=1/4(arcsinx)^2-1/4arcsinx*2sin(arcsinx)cos(arcsinx)-1/8[1-2sin^2(arcsinx)+C
=1/4(arcsinx)^2-1/2arcsinx*x*√(1-x^2)-1/8+1/4x^2+C

Answer 2

美女，祝你考试顺利，我只是路过！

Answer 3