Erm, I have this integration question:
sqrt((1-cos(x))/2) with an upper limit of 2pie and lower limit of 0.
I can't seem to do it right.
I tried simplifying it:
= 1 - (cos^-2(x))^(1/2)
= 1 - cos^-1(x)
Then I intgrated 1... which becomes x, but I have no idea how to do cos^-1(x), I saw the long formula for the integration of cos^something(x) but I don't think it applies here... because the answer is 4...
Does the above really say the following?
Originally Posted by Dorumonsg
If so, how did you get that? (Wondering)
a sqrt is something to the power of 1/2
x/something = x^-something?
sqrt((1-cos(x))/2) = sin(x/2)
That's your basic half angle formula from Trig
Note when you integrate sin(x/2) from 0 to 2pi you do get 4
Sin(x/2) after integrating I have -cos(x/2), I can't get 4? And anyone knows where I can get a complete list of trigo formulas? I never really learned them before?
after integrating you have -2cos(x/2) from 0 to 2 pi
you then have -2cos(pi) - (-2cos(0)) = -2(-1) + 2 = 4
you should have a list of trig formulas in your Calculus book--if not
get a trig book or Google Trig Formulas -- you should find them somewhere
How did you get -2cos(x/2)?
Originally Posted by Calculus26
If I take x/2 as U.
And differentiate it I get -2x^-3
So after integration I have 1/(-2x^-3)(-cos(x/2))?
if u = x/2 = (1/2) x
du = 1/2 dx
2du = dx
the 1/2 is a coefficient not a power--for some reason you're
reading x/2 as 1/(x^2) not (1/2) x
Oh okie, I get it. But just to double check if I have a number :
X^2, is it same as 1/X^-2
or X^-2 = 1/X^2
yes in that case you would be correct