Solving Difficult Integral Homework with Limits

[sara_87]

Homework Statement

I want to evaluate the integral:

\int \frac{\sqrt{a^2-x^2}(x-b)}{x^2-c^2}dx

Homework Equations

the integration limits are d (lower limit) and a (upper limit)

The Attempt at a Solution

First, I split the integral into 2 parts:

\int \frac{\sqrt{a^2-x^2}(x)}{x^2-c^2}dx- \int \frac{\sqrt{a^2-x^2}(b)}{x^2-c^2}dx<br />

Then, i decided to deal with each part separately.
For the first part, i tried to use a substitution: let u=x^2, this gives:
\int \frac{\sqrt{a^2-u}}{2(u-c^2)}du

But, i thave no idea how to continue from there. the second part i also find difficult.
Does anyone have any ideas on the substitution i should consider?

Thank you very much in advance.

 

[sara_87]

the denominator in the last integral should read:
2(u-c^2)

 

[dextercioby]

Well, making x=a\sin t [/tex] in the original integral throws away the square root. Then you will return to an integral of a rational function by making a new substitution: \tan\frac{t}{2} = u.

 

[sara_87]

After making the substitution x=a*sin(t), I get:

\int \frac{a^2 cos^2(t)(a sin(t)-b)}{a^2 sin^2(t)-c^2}dt

but if i then use the substitution: tan(t/2)=u,
this means: \frac{sec^2(t/2)}{2}dt=du
so,
dt=2cos^2(t/2)du

How does this reduce the problem?