Question

How do you integrate `int (6-x^2)/sqrt(9-x^2) dx` using trigonometric substitution?

Answer 1

`1/2xsqrt(9-x^2)+3/2arcsin(x/3)+C`

Explanation:

`I=int(6-x^2)/sqrt(9-x^2)dx`

Rewrite the integral for simplification:

`I=int(-3+9-x^2)/sqrt(9-x^2)dx=-3intdx/sqrt(9-x^2)+int(9-x^2)/sqrt(9-x^2)dx`

`color(white)I=-3intdx/sqrt(9-x^2)+intsqrt(9-x^2)dx`

First focusing on `J=intdx/sqrt(9-x^2)`. For this, let `x=3sintheta`. This implies that `dx=3costhetad theta`.

`J=intdx/sqrt(9-x^2)=int(3costhetad theta)/sqrt(9-9sin^2theta)=intcostheta/sqrt(1-sin^2theta)d theta=intd theta`

`color(white)J=theta=arcsin(x/3)`

Recall that the `theta=arcsin(x/3)` relationship comes from our earlier substitution `x=3sintheta`.

Now we can work on the next integral `K=intsqrt(9-x^2)dx`. We will use the same substitution as before, `x=3sinphi`, implying that `dx=3cosphidphi`.

`K=intsqrt(9-x^2)dx=intsqrt(9-9sin^2phi)(3cosphidphi)`

`color(white)K=9intsqrt(1-sin^2phi)(cosphi)dphi=9intcos^2phidphi`

Rewrite by solving for `cos^2x` in the identity `cos2x=2cos^2x-1`.

`K=9int(cos2phi+1)/2dphi=9/2intcos2phidphi+9/2intdphi`

For the first integral, we can use the substitution `u=2phi` so `du=2dphi` and `dphi=1/2du`. The second integral is a fundamental one:

`K=9/4intcosudu+9/2phi=9/4sinu+9/2phi=9/4sin2phi+9/2phi`

Using `sin2x=2sinxcosx`:

`K=9/2sinphicosphi+9/2phi`

Now to undo these substitutions, recall that `x=3sinphi`. Thus, `phi=arcsin(x/3)` and `sinphi=x/3`. Thus `cosphi=sqrt(1-sin^2phi)=sqrt(1-x^2/9)=1/3sqrt(9-x^2)`. So:

`K=9/2(x/3)(1/3sqrt(9-x^2))+9/2arcsin(x/3)`

`color(white)K=1/2xsqrt(9-x^2)+9/2arcsin(x/3)`

Returning to the original integral:

`I=-3J+K`

`color(white)I=-3arcsin(x/3)+[1/2xsqrt(9-x^2)+9/2arcsin(x/3)]`

`color(white)I=1/2xsqrt(9-x^2)+3/2arcsin(x/3)+C`