Simplifying Trig Identities

[mill]

Homework Statement

The length of the curve r(t) = cos^3(t)j+sin^3(t)k, 0 =< t <= pi/2
is

Homework Equations

AL in polar = ∫sqrt(r^2 + [dr/dθ]^2)

The Attempt at a Solution

I am having trouble simplifying the terms within the square root. What method should I use to deal with the pieces?

r^2 = (cos^3(t))^2 + 2 cost^3(t)sin^3(t) + (sin^3(t))^2

dr/dθ = 3(cos(t)^2)sin(t) + 3(sin(t)^2)cos(t)

[dr/dθ]^2 = 9[cos^4(t)sin^2(t) + 2cos^3(t)sin^3(t) + sin^4(t)cos^2(t))

sqrt(cos^3(t))^2 + 2 cost^3(t)sin^3(t) + (sin^3(t))^2 + 9[cos^4(t)sin^2(t) + 2cos^3(t)sin^3(t) + sin^4(t)cos^2(t))

Simplified a bit:

sqrt(1 + 4cos^3(t)sin^3(t) + 9 cos^4(t)sin^4(t) +9sin^4(t)cos^2(t))

How would I further simplify from this?

 

[haruspex]

There seem to be some inconsistencies in your notation.
From the equation for r(t), I guess j and k are unit vectors, so r(t) is a vector. But your equation for arc length treats r as a scalar. This seems to have led to an error here:
$r^2 = (\cos^3(t))^2 + 2 \cos^3(t)\sin^3(t) + (\sin^3(t))^2$

 

[mill]

There seem to be some inconsistencies in your notation.
From the equation for r(t), I guess j and k are unit vectors, so r(t) is a vector. But your equation for arc length treats r as a scalar. This seems to have led to an error here:
$r^2 = (\cos^3(t))^2 + 2 \cos^3(t)\sin^3(t) + (\sin^3(t))^2$

Oh, I see. Thanks.