How do you integrate #x^2/(sqrt(9-x^2))#?
1 Answer
Explanation:
#I=intx^2/sqrt(9-x^2)dx#
Rewrite
#I=int(x^2-9+9)/sqrt(9-x^2)dx=int(x^2-9)/sqrt(9-x^2)dx+int9/sqrt(9-x^2)dx#
Rewriting for simplification:
#I=-int(9-x^2)/sqrt(9-x^2)dx+int9/sqrt(9-x^2)dx#
#I=-intsqrt(9-x^2)dx+int9/sqrt(9-x^2)dx#
Let
#J=-intsqrt(9-x^2)dx#
Let
#J=-intsqrt(9-9sin^2theta)(3costhetad theta)#
#J=-3intsqrt9sqrt(1-sin^2theta)(costheta)d theta#
#J=-9intcos^2thetad theta#
Using
#J=-9int(cos2theta+1)/2d theta=-9/2intcos2thetad theta-9/2intd theta#
Solve the first integral by sight or by substitution:
#J=-9/4sin2theta-9/2theta#
Using
#J=-9/2sinthetacostheta-9/2theta#
From our substitution
Furthermore, we see that
#J=-9/2(x/3)(1/3sqrt(9-x^2))-9/2arcsin(x/3)#
#J=-1/2xsqrt(9-x^2)-9/2arcsin(x/3)#
Now solving for
#K=int9/sqrt(9-x^2)dx#
We will use the same substitution,
#K=int9/sqrt(9-9sin^2phi)(3cosphidphi)#
#K=int(27cosphi)/(sqrt9sqrt(1-sin^2phi))dphi#
#K=int(9cosphi)/cosphidphi=9intdphi=9phi#
From
#K=9arcsin(x/3)#
Now that we've solved for
#I=J+K#
#I=[-1/2xsqrt(9-x^2)-9/2arcsin(x/3)]+9arcsin(x/3)#
#I=-1/2xsqrt(9-x^2)+9/2arcsin(x/3)#
#I=(9arcsin(x/3)-xsqrt(9-x^2))/2+C#