Determine the area in the 1st quadrant bounded by $\displaystyle x=acos^3t$, $\displaystyle y=asin^3t$, the x-axis and the y-axis (a>0)

In my solution book it says:

As x varies from 0 to a, t varies from $\displaystyle \frac {\pi} {2}$ to 0

How do you know that?

Skipping all the way down to this step

$\displaystyle 3a^2\int_\frac{\pi} {2} ^2 sin^4tdt - 3a^2\int_\frac{\pi} {2} ^2 sin^6tdt$

= $\displaystyle [3a^2*(\frac{1} {2}*\frac{3} {4}*\frac{\pi} {2}] - [3a^2*(\frac{1} {2}*\frac{3} {4}*\frac{5} {6}*\frac{\pi} {2}]$

How do you get to that step?

Thanks a lot.