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?

Thank you for your help

 

[uart]

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?

Thank you for your help

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.

 

[uart]

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

 

[uart]

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

Ok so the question has now been changed to a definite integral, is that right? Sometimes there are more options available in how to handle a definite integral so it makes a difference (in general, though I'm not sure if it will make a difference this case, I'll take another look).

 

[pfsm01]

If the answer for the definite integral is in fact πb-π√(b^2-a^2 ) for limits x=-a to +a, would it be incorrect to back-calculate the answer for the indefinite integral? For instance, bπ((a+x)/2a) - π√((b^2)-(a^2))((a+x)/2a) with limits x=-a to +a resolves into the answer above.