Find the proper point on astroid for conditions to be met, using definite-integral for finding length. I was given the following task:
We have an astroid $x=acos^3(t), y=asin^3(t)$ and two points on it $A(a,0), B(0,a)$. We need to find point $M$, which belongs to the arc $AB$, and the length of arc $AM$ should be equal to $\frac{1}{4}$ length of arc $AB$.
As I understood I need to find such $x_0, y_0$ for point $M(x_0, y_0)$ to meet all the conditions. From the task we have $x(t), y(t)$, which means that we need to find the proper $t_0$ for $x(t_0)=x_0, y(t_0)=y_0$. Let the length of $AB$ equals $l$ and $\frac{1}{4}l=l_0$, then as follows from my definite-integrals course: $$l=\int_0 ^\frac{\pi}{2}\sqrt{[-3acos^2(t)sin(t)]^2+[3asin^2(t)cos(t)]^2}dt=$$$$3a\int_0 ^\frac{\pi}{2}\sqrt{cos^2(t)sin^2(t)[cos^2(t)+sin^2(t)]}dt=$$$$3a\int_0^\frac{\pi}{2}cos(t)sin (t)dt$$

Here I make a substitution $u=sin(t)$, then $du=cos(t)dt, u_1=sin(0)=0, u_2=sin(\frac{\pi}{2})=1$, so now we have:
$$3a\int_0 ^1 udu=\frac{3au^2}{2} \Biggr|_0^1=\frac{3a}{2}$$
First question. Is the integrand solved correctly?
Second question. What should I do next? 
P.S. Right now we are dealing with definite-integrals on our math course, so it is supposed that I use definite-integral in this task.

Hint:
Set $\displaystyle 3a\int_0^{t_1}\sin t\cos t\;dt=\frac{1}{4}\cdot 3a\int_0^{\frac{\pi}{2}}\sin t\cos t\;dt$,  and then solve for $t_1$.
