### Calculus Forgotten

Today over lunch my friend and I challenged each other with math questions. Below is the one he gave me. To my embarassment, I had forgotten how to even begin to solve it...

Well, do

Well, do

*you*remember?
## 3 Comments:

i'm kinda rusty, but i think you need to solve that via trig substitution.

∫ (1-x^2)^1/2 dx

assume x = sin k; then dx = cos k dk ->

∫ (1-(sin k)^2)^1/2 (cos k dk) ->

∫ (1-sin^2 k)^1/2 (cos k dk) ->

∫ (cos^2 k)^1/2 (cos k dk) ->

∫ cos k (cos k dk) ->

∫ cos^2 k dk

...

i think you should end up with 1/2(x(1-x^2)^(1/2) + sin^-1 x) + c

By w, at 8/21/2010 10:35 PM

The primitive is:

F(x) = 1/2 * (sqrt(1-x^2)*x + arcsin(x))

So F(x)|(0,1) = F(1) - F(0) = pi / 4

By Alex Ksikes, at 10/27/2010 2:10 AM

@w, @Alex - Well guys, your calculus is better than mine, I must say.

By Jonathan, at 7/21/2011 8:09 PM

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