How do you integrate #x^2/(sqrt(9-x^2))#?

Question

Emilia Aguilar · Accepted Answer

#(9arcsin(x/3)-xsqrt(9-x^2))/2+C#

#I=intx^2/sqrt(9-x^2)dx#

Rewrite #x^2# as #x^2-9+9#:

#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# and #K=int9/sqrt(9-x^2)dx#.

#J=-intsqrt(9-x^2)dx#

Let #x=3sintheta# so that #dx=3costhetad theta#:

#J=-intsqrt(9-9sin^2theta)(3costhetad theta)#

#J=-3intsqrt9sqrt(1-sin^2theta)(costheta)d theta#

#J=-9intcos^2thetad theta#

Using #cos2theta=2cos^2theta-1# so solve for #cos^2theta#:

#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 #sin2theta=2sinthetacostheta#:

#J=-9/2sinthetacostheta-9/2theta#

From our substitution #x=3sintheta# we see that #theta=arcsin(x/3)#.

Furthermore, we see that #sintheta=x/3# and #costheta=sqrt(1-sin^2theta)=sqrt(1-x^2/9)=1/3sqrt(9-x^2)#. Thus:

#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#:

#K=int9/sqrt(9-x^2)dx#

We will use the same substitution, #x=3sinphi#, so #dx=3cosphidphi#.

#K=int9/sqrt(9-9sin^2phi)(3cosphidphi)#

#K=int(27cosphi)/(sqrt9sqrt(1-sin^2phi))dphi#

#K=int(9cosphi)/cosphidphi=9intdphi=9phi#

From #x=3sinphi# we see that #phi=arcsin(x/3)#:

#K=9arcsin(x/3)#

Now that we've solved for #J# and #K#, return to #I#:

#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#

Axel Alvarez · Answer

undefined To integrate $ \frac{x^2}{\sqrt{9-x^2}} $, you can use trigonometric substitution. Let $ x = 3\sin(\theta) $, then $ dx = 3\cos(\theta) d\theta $. Substitute these expressions into the integral, rewrite $ \sqrt{9-x^2} $ as $ 3\cos(\theta) $, and simplify the integrand. The resulting integral will involve trigonometric functions, which can then be integrated using standard trigonometric identities and techniques.