# Trigo Integration

• Nov 13th 2009, 08:22 AM
Dorumonsg
Trigo Integration
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:

sqrt((1-cos(x))/2)
= ((1-cos(x))/2)^(1/2)
= 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...
• Nov 13th 2009, 08:28 AM
stapel
Quote:

Originally Posted by Dorumonsg
...= ((1-cos(x))/2)^(1/2)
= 1 - (cos^-2(x))^(1/2)

Does the above really say the following?

$\sqrt{\frac{1\, -\, \cos(x)}{2}}\, =\, 1\, -\, \sqrt{\frac{1}{\cos^2(x)}}$

If so, how did you get that? (Wondering)
• Nov 13th 2009, 08:31 AM
Dorumonsg
a sqrt is something to the power of 1/2

and

x/something = x^-something?
• Nov 13th 2009, 08:36 AM
Calculus26
Use

sqrt((1-cos(x))/2) = sin(x/2)
• Nov 13th 2009, 08:38 AM
Dorumonsg
How did u get tat?
• Nov 13th 2009, 08:40 AM
Calculus26
That's your basic half angle formula from Trig

Note when you integrate sin(x/2) from 0 to 2pi you do get 4
• Nov 13th 2009, 08:56 AM
Dorumonsg
Erm...

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?
• Nov 13th 2009, 09:01 AM
Calculus26
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
• Nov 13th 2009, 09:07 AM
Dorumonsg
Quote:

Originally Posted by Calculus26
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)?

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))?
• Nov 13th 2009, 09:11 AM
Calculus26
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
• Nov 13th 2009, 09:24 AM
Dorumonsg
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
• Nov 13th 2009, 09:32 AM
Calculus26
yes in that case you would be correct