Question

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

Answer 1

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

Explanation:

`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`