Question

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

Answer 1

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