Question

根号下4-x^2的原函数是什么，就是定积分

Answer 1

设x=2sint  ，

Answer 2

F(x)=∫√(4-x^2)dx=2∫√[1-(x/2)^2]dx
|x/2|≤1，
令sint=x/2，则x=2sint
F(x)=2∫√(1-sint^2)d(2sint)=4∫cost^2dt
=4∫costd(sint)=4costsint-4∫sintd(cost)=4costsint+4∫sint^2dt
=4costsint+4∫(1-cost^2)dt=4costsint+4t-4∫cost^2dt
=4costsint+4t-F(x)
F(x)=2costsint+2t+C=sin2t+2t+C
=sin[2arcsin(x/2)]+2arcsin(x/2)+C

Answer 3

评一下吧

Answer 4

令：x=2sint ，dx=2costdt ,√（4-x^2) =2cost
∫ √（4-x^2) dx
=∫2cost*2cost dt
= 2∫(1+cos2t)dt
= 2[t-1/2*sin2t)] + C
= 2[t-sintcost)] + C
= 2[arcsin(x/2) - x/2 *1/2*√（4-x^2)] + C
= 2*arcsin(x/2) - 1/2*x*√（4-x^2)] + C

Answer 5

采用变量置换法解上式即设∫√(2^2-x^2)dx其中设x=2sint,dx=2costdt代入则可.
脑筋应让提问人自己动!