Expert: Paul Klarreich - 2/2/2006

Question
So to put it as a +/- answer, I would have to write it as:
- arcsin(x/3)
or would it be the entire
 - sqr(9-x^2)
= ------------ - arcsin(x/3)
     x
Sorry for being kind of stupid; it's been a LONG time since I've had to do these sorts of things. Thanks again!
-------------------------
Followup To
Question -
I have been stumped on this problem and finally have given up. Would you care to help me?
It is:
S(square root of 9 minus X squared), over x squared, dx.
If you need me to, I can save the problem as an image. Thanks much in advance for any help!
Answer -
Hi, Ben,
Your Question:  I have been stumped on this problem and finally have given up.

Would you care to help me?
It is:
S(square root of 9 minus X squared), over x squared, dx.
If you need me to, I can save the problem as an image. Thanks much in advance for any help!

------------------
WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

Is this it?  If so, you can find a formula for it at:
http://www.sosmath.com/tables/integral/integ13/integ13.html

{ sqrt(9-x^2)
| ----------- dx
}     x^2

But of course, you want to know how to do these.  So my first inclination would be to do 'trig-substitution.'  If you never heard of these, of course, you are in trouble, but here is how they work:

Look at the radical,  sqrt(9-x^2), as a side of a right triangle.  Since you have  3^2 - x^2, it is natural to make '3' the hypotenuse.  Call one of the angles theta (I will write t for theta) and put the x opposite the theta.  It looks like this.  You will see, of course, that the third side is the square root radical in your problem.

  /|
3/ |
/  | x    <<-- crude attempt at a diagram
/___|
radical

        x
sin t = ---,  or  x = 3 sin t
        3
       radical
cos t = ------- or  sqrt(...) = 3 cos t
         3

Since  x = 3 sin t,  dx = 3 cos t dt

Now you can do some substitutions:

{ sqrt(9-x^2)
| ----------- dx =
}     x^2

{  3 cos t
| ----------- 3 cos t dt =
}  9 sin^2(t)

{  cos^2(t)
| ----------- dt = and now you use some trig stuff:
}  sin^2(t)   

{ (1 - sin^2(t))
| -------------- dt =
}  sin^2(t)   

{
| (csc^2(t) - 1) dt = - cot t - t
}

[We are all expected to remember that D(cot x) = - csc^2(x)]

Now you have to substitute back. You get the expression for the cotangent by looking at the triangle and picking the correct sides.

 - sqr(9-x^2)
= ------------ - arcsin(x/3)
     x


Answer
Hi, Ben,

Your question:
So to put it as a +/- answer, I would have to write it as:
- arcsin(x/3)
or would it be the entire
 - sqr(9-x^2)
= ------------ - arcsin(x/3)
     x
--------------------------------
Yes, yes, the whole thing.  If you are not totally convinced, all you have to do is check.  Remember, you check an integration exercise by differentiating the answer.  You will need the quotient rule for the first term and the chain rule for the second, of course, and then a bit of simplification.