题目

∫x^3*√(a^2-x^2)dx (a>0,为常数) ∫x^2/√(9-x^2)dx

提问时间：2021-03-03

答案

t=√(a^2-x^2),x^2=a^2-t^2
Sx^3*√(a^2-x^2)dx=-1/2*Sx^2*√(a^2-x^2)d(a^2-x^2)
=-1/2*S(a^2-t^2)tdt^2
=-S(a^2-t^2)t^2dt
=S(t^4-a^2 *t^2)dt
=1/5*t^5-a^2/3*t^3+c
=1/5*(√(a^2-x^2)^5-a^2/3*(√(a^2-x^2)3+c
x=3sint,t=arctanx/3,dx=3costdt
∫x^2/√(9-x^2)dx
=S9(sint)^2*3cost*3cost*dt
=81/4*S(sin2t)^2dt
=81/32*S(1-cos4t)d4t
=81/32*4t-81/32*sin4t+c
,t=arctanx/3代入化简即可