Question

定积分问题?

Answer 1

let
x= asinu
dx=acosu du
x=0, u=0
x=a, u=π/2
∫(0->a) x^2.√(a^2-x^2) dx
=a^4.∫(0->π/2) (sinu.cosu)^2 du
=(1/4)a^4.∫(0->π/2) (sin2u)^2 du
=(1/8)a^4.∫(0->π/2) (1-cos4u) du
=(1/8)a^4 .[u-(1/4)sin4u]|(0->π/2)
=(1/16)a^4.π

Answer 2

x=asint带人很快就得到被积函数=(asint)^2 acost dasint = a^4 (sintcost)^2

=a^4(sin2t)^2 /4 = a^4 (1-cos2t)/8
积分=a^4 t/8 - a^4 sin2t/16+C

然后把x=arcsint带人即可

Answer 3

详细过程如图……所示