Having trouble solving question involving parametric equations I have been given the following:
$$y = a \cdot \cos^3t$$  $$x = a \cdot \sin^3t$$  
$$0 \leqslant t \leqslant {\frac\pi2}$$
I am supposed to show that the mean value of $y$ over the interval $0\leqslant x \leqslant a$ is given by:  
$$m = 3a\int^{\frac12\pi}_0(\cos^4t-\cos^6t)dt$$
I know that the mean fomula is given by:  
$$m=\frac1{b-a}\int^b_aydx$$
But I'm having trouble finding an equation for $y$ in terms of $x$.
In addition, I am confused as to how the final answer will be given as an integral in terms of $t$.
I thought I should differentiate $y$, $x$ in terms of $t$ then maybe I would find a substitution:  
$$\frac{dy}{dt} = -3a \cdot \cos^2t \cdot \sin t$$
$$\frac{dx}{dt} = 3a \cdot \cos t \cdot \sin^2t$$
$$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{-3a \cdot \cos^2t \cdot \sin t}{3a \cdot \cos t \cdot \sin^2t} = -\frac{\cos t}{\sin t} = -\cot t$$
Now I'm facing two problems:  


*

*I can't integrate both sides to get $y$ in terms of $x$ because the trigonometric function on R.H.S. is still in terms of $t$  

*Even if I did integrate this method seems too inefficient


What's the most appropriate method of solving this?

Think of the parametric equations as a change of variable in the integral defining $m$:
$$
m = \frac{1}{a} \int_0^a y \, dx = \frac{1}{a} \int_0^{\pi/2} y \frac{dx}{dt} \, dt
$$
Now plug in the equations for $x$ and $y$ in terms of $t$ to get the desired result:
\begin{align}
m &= \frac{1}{a} \int_0^{\pi/2} a \cos^3 t \left(3a \cos t (1 - \cos^2 t )\right) \, dt \\
&= 3a \int_0^{\pi/2} \left(\cos^4 t - \cos^6 t\right) \,dt
\end{align}
