题目内容：

根号（9-x^2）在[0,2]上的定积分怎么求,知道的朋友可以告诉我吗,

优质解答

令x=3sint则原式=∫(0→arcsin(2/3))3cost*3costdt=9∫(0→arcsin(2/3))cos^2(t)dt=9/2∫(0→arcsin(2/3))(cos(2t)+1)dt=9/4sin(2t)|(0→arcsin(2/3))+9/2t|(0→arcsin(2/3))=9/2sintcost|(0→arcsin(2/3))+9/2t|(0...