Homework Help: Help with evaluating indefinite integral

pfsm01 · Sep 20, 2012

1. The problem statement, all variables and given/known data

Evaluate the integral:

∫ {√[(a^2)-(x^2)] / (b-x)} dx

2. Relevant equations

∫ u dv = uv - ∫v du

3. 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 · Sep 20, 2012

Use trig substitution. Set x = asin(theta) and sub in x and dx.

pfsm01 · Sep 20, 2012

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 · Sep 21, 2012

pfsm01 said: ↑

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 [itex]x = a \sin(\theta)[/itex] 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:
[tex]\int \frac{a^2 \cos^2 \theta}{b - \sin \theta}\, d\theta[/tex]
which doesn't look too easy.

uart · Sep 21, 2012

pfsm01 said: ↑

1. The problem statement, all variables and given/known data

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.

[tex]\frac{\sqrt{a^2 - x^2}}{a-x} = \frac{\sqrt{a+x}}{\sqrt{a-x}} = \frac{a+x}{\sqrt{a^2-x^2}}[/tex]

So for this special case we have,
[tex] \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[/tex]
which are both very easy integrals. For the general case however, hmmm, hopefully someone else can help out.

pfsm01 · Sep 21, 2012

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

uart · Sep 21, 2012

pfsm01 said: ↑

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 · Sep 24, 2012

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.