第1个回答 2011-11-28
①∫dx/(a²-x²)^(3/2)=∫dx/[(a²-x²)√(a²-x²)]=∫dx/{(a²-x²)a√[1-(x/a)²]}
令x/a=sinu,则x=asinu,dx=acosudu,代入上式得:
=∫acosudu/[a²-a²sin²u)a√(1-sin²u)
=∫du/(a²cos²u)=(1/a²)tanu+C=(1/a²)[x/√(a²-x²)]+C
②
4-x²=2²-x²,令x=2sinu则dx=2cosudu,sinu=x/2
√(4-x²)=√(4-4sin²u)=2cosu,cosu=(1/2)√(4-x²)
∴∫(x-2)√(4-x²) dx
=∫(2sinu-2)2cosu*2cosu du
=8∫(sinu-1)cos²u du
=8∫sinucos²u du-8∫cos²u du
=-8∫cos²u d(cosu)-4∫(1+cos2u) du
=(-8/3)cos³u-4(u+1/2*sin2u)+C
=(-8/3)[1/2*√(4-x²)]³-4arcsin(x/2)-4*x/2*(1/2)√(4-x²)+C
=(-1/3)(4-x²)^(3/2)-4arcsin(x/2)-x√(4-x²)+C