No.

In polar coordinates,

L = sqrt[r^2 +(dr/dT)^2]dT -------***

It is not

L = sqrt[1 +(dr/dT)^2]dT

You are confused with

L = sqrt[1 +(dy/dx)^2]dx

in cartesian coordinates.

------

r = 1 +cosT

dr/dT = -sinT

If the limits are from T=0 to T=2 radians,

L = INT.(0->2)[sqrt{(1+cosT)^2 +(-sinT)^2}]dT

L = INT.(0->2)[sqrt{(1 +2cosT +cos^2(T)) +sin^2(T)}]dT

L = INT.(0->2)[sqrt{(1 +2cosT +{cos^2(T) +sin^2(T)}}]dT

L = INT.(0->2)[sqrt{1 +2cosT +1}]dT

L = INT.(0->2)[sqrt{2 +2cosT}]dT ---(1)

The "dT" outside of the bracket is not the derivative of the sqrt{2 +2cosT} inside the bracket.

We need to express the sqrt{2 +2cosT} into a one-degree function.

See if we can convert the (2 +2cosT) into a square of cos(nT) or sin(nT).

cosT

= cos(T/2 +T/2)

= cos^2(T/2) -sin^2(T/2)

= cos^2(T/2) -[1 -cos^2(T/2)]

= 2cos^2(T/2) -1 ---***

Substitute that into (1),

L = INT.(0->2)[sqrt{2 +2cosT}]dT ---(1)

L = INT.(0->2)[sqrt{2 +2(2cos^2(T/2) -1)}]dT

L = INT.(0->2)[sqrt{2 +4cos^2(T/2) -2)}]dT

L = INT.(0->2)[sqrt{4cos^2(T/2)}]dT

L = INT.(0->2)[2cos(T/2)]dT

Now we manipulate the "dT" outside.

L = INT.(0->2)[2cos(T/2)](2dT/2)

L = INT.(0->2)[2cos(T/2)](dT/2)(2)

L = 2*2*INT.(0->2)[cos(T/2)](dT/2)

L = 4*INT.(0->2)[cos(T/2)](dT/2)

Now we can integrate,

L = 4[sin(T/2)](0->2)

L = 4[sin(2/2) -sin(0/2)]

L = 4[sin(1) -sin(0)]

L = 4[0.84147 -0]

L = 3.366 ----answer.