一道定积分求根号（9-x^2）分之x^2的定积分

Question

一道定积分
求根号（9-x^2）分之x^2的定积分

Answer 1

没有x的取值区间,故为不定积分.
令x=3sint,则dx=3cotdt
原式=∫√{(3sint)^2/[9-(3sint)^2]^(1/2)}*3costdt
=∫9(sint)^2dt
=9/2*∫(1-cos2t)dt
=9/2*(t-1/2sin2t)+C
=9/2*[arcsin(x/3)-1/2sin2(arcsin(x/3)]+C
=9/2*arcsin(x/3)-9/4*2x/3*(1-x^2/9)^(1/2)+C
=9/2*arcsin(x/3)-3x/2*(1-x^2/9)^(1/2)+C