Integrate sqrt((1-cos(x))/2) from 0 to 2π

In summary, the conversation discusses an integration question involving the expression sqrt((1-cos(x))/2) with upper and lower limits. The individual has attempted to simplify the expression but is unsure how to integrate cos^-1(x). It is suggested to look at trigonometric substitutions in notes or textbooks to make the integrand easier. There is also a reminder that sqrt(a + b) is not equal to a + sqrt(b).
  • #1
DorumonSg
64
0
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...
 
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  • #2
DorumonSg said:
sqrt((1-cos(x))/2)
= ((1-cos(x))/2)^(1/2)
= 1 - (cos^-2(x))^(1/2)

bad.. BAD Dorumon! =P

sqrt(a + b) =/= a + sqrt(b)

You should have some trigonometric substitutions in your notes/textbook that involve cosine. Take a look at them, and see if they make the integrand a little nicer.
 

FAQ: Integrate sqrt((1-cos(x))/2) from 0 to 2π

What is the formula for integrating sqrt((1-cos(x))/2) from 0 to 2π?

The formula for integrating this function is ∫√((1-cos(x))/2) dx from 0 to 2π.

What is the significance of integrating this particular function?

This function represents the area under the curve of a semi-circle with a radius of 1, which has many applications in physics and engineering.

Can this integral be solved analytically?

Yes, this integral can be solved analytically using trigonometric identities and integration techniques.

What are the limits of integration for this integral?

The limits of integration for this integral are 0 and 2π, representing the full range of values for the function.

How is this integral typically used in real-world applications?

This integral is often used to calculate the area of a half-circle or to find the work done by a force that varies in a circular motion.

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