Find the arc length of the parametric curve.

[hank]

x = cos^3 t y = sin^3 t and z = 2; 0 <= t <= pi/2

r'(t) = < -3sin^2 t, 3cos^2 t, 0>

||r'(t)|| = sqrt[(-3sin^2 t)^2 + (3cos^2 t)^2 + 0^2]
||r'(t)|| = sqrt[9sin^4 t + 9cos^4 t]
||r'(t)|| = 3 * sqrt[sin^4 t + cos^4 t]

And, this is where I get stuck.
I suspect that sin^4 t + cos^4 t can be reduced to some nice square rootable function, but I can't see it.
Is there something else I can do here? Help?

 

[galactus]

$\displaystyle \frac{dx}{dt}=-3sin(t)cos^{2}(t), \;\ \frac{dy}{dt}=3sin^{2}(t)cos(t), \;\ \frac{dz}{dt}=0$

$\displaystyle \sqrt{9sin^{2}(t)cos^{4}(t)+9sin^{4}(t)cos^{2}(t)}=3\sqrt{sin^{2}(t)cos^{2}(t)\underbrace{(cos^{2}(t)+sin^{2}(t)}_{\text{equals 1}})}$

$\displaystyle =\int{3sin(t)cos(t)}dt$

But, for your info $\displaystyle sin^{4}(t)+cos^{4}(t)=\frac{1}{4}(cos(4t)+3)$