Question

What is the integral of sqrt(9-x^2)/x?

Answer 1

`intsqrt(9-x^2)/xdx=ln|(3-sqrt(9-x^2))/3|+sqrt(9-x^2)+C`

Explanation:

We want `intsqrt(9-x^2)/xdx`.

In general, when encountering an integral involving `sqrt(a^2-x^2),` make the trigonometric substitution `x=asintheta.`

In this case, `a^2=9, a=3,` and so our substitution is `x=3sintheta`.

Solving for `dx` yields

`dx=3costhetad theta`.

Apply the substitution:

`int(cancel3costhetasqrt(9-9sin^2theta))/(cancel3sintheta)d theta`

Simplify:

`int(costhetasqrt(9(1-sin^2theta)))/sinthetad theta`

Recall the identity

`sin^2theta+cos^2theta=1`

This identity also tells us that

`1-sin^2theta=cos^2theta`

Rewrite with the identity applied:

`3int(costhetasqrt(cos^2theta))/sinthetad theta=3int(cos^2theta/sintheta)d theta`

Now, to solve the resultant integral, we're best off rewriting again with the identity reversed:

`3int(1-sin^2theta)/sinthetad theta=3(intcscthetad theta-intsinthetad theta)`

`intcscthetad theta=ln|csctheta-cottheta|`. Memorize this, it's a fairly common integral.

`intsinthetad theta=-costheta`

`3ln|csctheta-cottheta|+3costheta+C` (Don't forget the constant of integration).

We now need to rewrite in terms of `x.` We said `x=3sintheta,` so `sintheta=x/3, csctheta=1/(x/3)=3/x.`

To solve for cosine, recall `sin^2theta+cos^2theta=1, x^2/9+cos^2theta=9/9, cos^2theta=(9-x^2)/9, costheta=sqrt(9-x^2)/3.`

Finally, we end up with

`intsqrt(9-x^2)/xdx=ln|(3-sqrt(9-x^2))/3|+sqrt(9-x^2)+C`

Answer 2

`I = sqrt(9 - x^2)/3 - ln|(3 + sqrt(9 - x^2))/x| + C`

Explanation:

I would solve this using a trig substitution. Let `x = 3sintheta`. Then `dx = 3costheta d theta`.

`I = int sqrt(9 - 9sin^2theta)/(3sintheta) * 3costheta d theta`

`I = int sqrt(9cos^2theta)/sintheta * costheta d theta`

`I = int 3cos^2theta/sintheta d theta`

`I = 3int (1 - sin^2theta)/sintheta d theta`

`I = 3int csctheta - sinthetad theta`

These are two known integrals.

`I = costheta - ln|csctheta + cottheta| + C`

Recall from the initial substitution that `x/3 =sintheta`, therefore `sqrt(9 - x^2)/3 = costheta`, `3/x = csctheta` and `sqrt(9 - x^2)/x = cottehta`.

`I = sqrt(9 - x^2)/3 - ln|3/x + sqrt(9 - x^2)/x| + C`

`I = sqrt(9 - x^2)/3 - ln|(3 + sqrt(9 - x^2))/x| + C`

Hopefully this helps!