It is easy to obeserve that is...
Kind regards
This one has had me stuck for a good hr.
The integral has a min of 0 and a max of 1.
S (4-x)/(sqrt(4-x^2)) dx
I'm trying to use the formula S (du)/(sqrt(a^2-u^2)) = arcsin u/a + C
I set "u" = (x^2), "du" = 2x, and a = 2.
Typically the problems I get will have the du equal the numerator and it's pretty easy from there. But my du does not equal the numerator and I'm stuck.
I think the answer for the interval (disregarding the min and max) is:
Sqrt(4-x^2) + 4 arcsin(x/2) + C.
And the part in underlined bold I have absolutely no idea where it came from.
And the final answer:
-2 + Sqrt(3) + (2pi)/3
Basically I cannot figure a way to get that answer or the steps involved. Can anyone help? Thank you.
I am still not following.
@Chisigma. I see that you broke them up to two separate integrals and divided by 2. But after that I don't see where to go from there.
@TheCoffeeMachine. I see what you did on the first two steps but on the 3rd one when you add them, I am lost. In the front of my calc book I have a sheet with basic integration formulas, it doesn't show anything that relates to yours. I also don't understand where you got the second form. Is their a good site with more formulas?
I also tried to look at how Wolfram Alpha solved it
(4-x)/(sqrt(4-x^2)) - Wolfram|Alpha
And the part that me throws me off on their solution is that they set x = 2sin(u) and u = arcsin (x/2)
I got a good 2+hrs into this one problem and 3 pages of scratch paper. I'm at a loss. Thank you in advance!
Derek Z
Just in case a picture helps...
... where (key in spoiler) ...
Spoiler:
Then back substitute from sin theta to x/2 to find F.
What TCM was hinting towards is kind of like this...
... where (key in spoiler) ...
Spoiler:
Edit: BTW, using the trig sub for the whole fraction like Wolfram does...
... means that after back-substituting you have an expression
which with some Pythag will iron out to the same as above.
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