how do i integrate this?

Question

i have to find the integral

S dx / (a^2 - x^2)^1/2

= S 1 / (a^2 - x^2)^1/2 dx

= S 1 / ((a+x)(a-x)^1/2) dx

this is as far as i have managed to take it.
where do i go from here?

Answer 1

You can make it into an arccos or arcsin (both of those work). Use the substitution:

x = a*cos(u) dx = -a*sin(u) du

After that you should be able to get it.

The answer will be arccos(x/a) + const

(or arcsin(x/a) + const, which is really just shifted by pi/2)

Answer 2

Your starting it incorrectly. It is natural to want to try to factor that demoniator, but this is a job for...
trig substitution!!!
This is actually a very easy problem if you know trig substitution or a very difficult and undoable problem if you don't.

For a moment let's declare x to be asin(t)

x = asin(t)
then dx = acos(t)dt
Now you have...
S 1/(a^2 - a^2sin^2(t))^1/2 (a cos(t) dt)
Factor the a^2 out in the denominator.
= S 1/[a^2(1 - sin^2(t))]^1/2 (a cos(t) dt)
Since sin^2(t) + cos^2(t) = 1 then 1 - sin^2(t) = cos^2(t)
This becomes...
S 1/(a^2 cos^2(t))^1/2 (a cos(t) dt)
The denominator is a perfect square! Take the sqaure root.
= S 1/(acos(t)) * (a cos(t) dt)
Everything cancels!!!! Now integrate 1 with respect to t.
= S dt
= t + c
Now go back and solve for t.
Since x = asin(t) then...
t = arcsin(x/a)
Thus...
S dx / (a^2 - x^2)^1/2 = arcsin(x/a) + c