astroid: vectors, curvature, arc length

Hope someone can help me out here, to check the work I've done and set me on the right path of the work I'm stuck at (or getting the wrong answer for!)

Astroid is defined by

$\displaystyle r(t) = (x(t), y(t) ) = (4cos^3t, 4sin^3t)$

a) Determine where r(t) defines a smooth curve, and find T(t) and N(t).

No idea about the first part, and here's what I get for $\displaystyle T(t):$

$\displaystyle T(t) = \frac{r'(t)}{||r'(t)||}$

$\displaystyle r'(t) = (-12sin(t)cos^2(t), 12sin^2(t)cos(t))$

$\displaystyle ||r'(t)|| = \sqrt{(-12sintcos^2t)^2 + (12sin^2tcost)^2} = 12sin(t)cos(t) or 6sin(2t) if you prefer$

$\displaystyle T(t) = \frac{(-12sin(t)cos^2(t), 12sin^2(t)cos(t))}{12sin(t)cos(t)}$

$\displaystyle T(t) = (-cos(t),sin(t))$

(should I put it in the form $\displaystyle T(t) = (-cost)i + (sint)j$?)

For $\displaystyle N(t)$ I got this:

$\displaystyle N(t) = \frac{T'(t)}{||T'(t)||}$

$\displaystyle T'(t) = (-sin(t), -cost(t))$

$\displaystyle ||T'(t)|| = \sqrt{(-sint)^2+(-cost)^2}=1$

$\displaystyle N(t) = (-sin(t), -cost(t))$

b. Find the curvature $\displaystyle \kappa(t)$ for $\displaystyle 0<t< \frac{\pi}{2} $

I used $\displaystyle \kappa(t) = \frac{||T'(t)||}{||r'(t)||}$

$\displaystyle \kappa(t) = \frac{1}{12sin(t)cost(t)}$

but $\displaystyle cos(\frac{\pi}{2}) = 0$, which means I'm dividing by zero and thus getting the wrong answer! Where have I gone wrong?

c. Determine the arc length s(t) for a section of the astroid starting at r(0) and ending at r(t) where $\displaystyle 0<t\leq \frac{\pi}{2}$

This is a quarter of the length round the astroid, so we can use the formula for the total length and divide by 4. The total length for an astroid $\displaystyle r(t) = (cos^3(t), sin^3(t))$ is 6, so the total length for this astroid will be 24, and a quarter length 6.

$\displaystyle s(t) = \int^{2\pi}_{0} r'(t)$

but here I get stuck! I know the answer comes to:

$\displaystyle s(t) = \int 6sin(2t)$

but integrating that I get -3*cos(2t) and both $\displaystyle cos(4\pi)$ and $\displaystyle cos(0) = 1$, which gives the answer of zero and not, as expected, 6.

as always, many thanks for any help given.