Evalute the integral from 0 to 2πa of (sqrrt of (a^2((1-cosx)^2+(sinx)^2)dx)?

Answer 1

expand the (1-cosx)^2 to get
1 -2cosx + (cosx)^2
then you have
sqrt( a^2(1 - 2cosx + (cosx)^2 + (sinx) ^2))

Use the trig identity (cosx)^2 + (sinx)^2 = 1
sqrt( a^2 ( 2 - 2cosx))

factor out a^2 and 4, you'll see why 4 soon
assuming a is positive.
2a * sqrt ( (1-cosx)/2 )

now you can use another trig identity
(1-cosx)/2 = (sin (x/2))2

2a * sqrt ( (sin(x/2))^2)
now you have a square root of a square

2a * | sin(x/2) |

I'll let you finish it from there :-)

Answer 2

the integral
={a^2[1+cos^2x-2cosx+sin^2x]}^1/2dx
I=(a^2-a^2cox)^1/2dx
=[a^2(1-cosx)]^1/2dx
=[a^2*2sin^2x/2)]^1/2dx
rt2a*(-2cos(x/2)
=-2rt2acos(x/2)
now apply the limits

Answer 3

http://integrals.wolfram.com/index.jsp