题目

∫（上限1,下限-1）(2-x^2)^(3/2) dx

提问时间：2021-02-10

答案

积分:(2-x^2)^(3/2)dx
=x(2-x^2)^(3/2)-积分:xd(2-x^2)^(3/2)
=x(2-x^2)^(3/2)-积分:x(-2x)*3/2*(2-x^2)^(1/2)dx
=x(2-x^2)^(3/2)+3积分:x^2(2-x^2)^(1/2)dx
现在求:积分:x^2(2-x^2)^(1/2)dx
令:x=根号(2)*cost,t[0,pi/2]
t=arccos(根号(2)*x/2)
所以:
积分:x^2(2-x^2)^(1/2)dx
=积分:2(cost)^2*根号(2)sint*根号(2)*(-sint)dt
=-4积分:(sintcost)^2dt
=-积分:(sin(2t))^2dt
=积分:(cos(4t)-1)dt
=1/4sin4t-t+C
x[0,1]
t[pi/2,pi/4]
该定积分为:
pi/4
积分上下限是:(0,1)
定积分为:
=x(2-x^2)^(3/2)+3积分:x^2(2-x^2)^(1/2)dx
=x(2-x^2)^(3/2)|(0,1)+3*pi/4
=1+3pi/4
因为被积函数是偶函数,所以最后的定积分为:
I =2*(1+3pi/4)
=2+3pi/3