# Help with evaluating indefinite integral

pfsm01

## Homework Statement

Evaluate the integral:

∫ {√[(a^2)-(x^2)] / (b-x)} dx

## Homework Equations

∫ u dv = uv - ∫v du

## The Attempt at a Solution

I've tried using integration by parts but it makes the integral even more complex.
I also tried using the table of integrals to find a solution to no avail.

Can someone please point me in the right direction?

Any help is much appreciated.

e^(i Pi)+1=0
Use trig substitution. Set x = asin(theta) and sub in x and dx.

pfsm01
Would I then integrate with respect to dtheta?

If I had to evaluate the integral from -a to +a, would I integrate from 0 to theta after the substitution?

Would I then integrate with respect to dtheta?

If I had to evaluate the integral from -a to +a, would I integrate from 0 to theta after the substitution?

If you were to substitute $x = a \sin(\theta)$ then the limits x=-a to +a would change to theta = -pi/2 to pi/2. However I don't think that substitution will help all that much? It changes it into:
$$\int \frac{a^2 \cos^2 \theta}{b - \sin \theta}\, d\theta$$
which doesn't look too easy.

## Homework Statement

Evaluate the integral:

∫ {√[(a^2)-(x^2)] / (b-x)} dx
That looks like a tricky integral pfsm. In the special case where b=a it simplifies (algebraically) to a very easy integral. For the general case however I'm stumped. Maybe if I show the simplified case it will give someone an else a lead.

$$\frac{\sqrt{a^2 - x^2}}{a-x} = \frac{\sqrt{a+x}}{\sqrt{a-x}} = \frac{a+x}{\sqrt{a^2-x^2}}$$

So for this special case we have,
$$\int \frac{\sqrt{a^2 - x^2}}{a-x} \, dx = \int \frac{a}{\sqrt{a^2-x^2}} \, dx + \int \frac{x}{\sqrt{a^2-x^2}} \, dx$$
which are both very easy integrals. For the general case however, hmmm, hopefully someone else can help out.

pfsm01
Apparently the answer is πb-π√(b^2-a^2 ) for limits x=-a to +a
but not sure how to arrive here