So, I know that $sin^2(t)$ is an even function so the $b_{n}$ co-efficient's are 0 and that we will only be figuring out the $a_{n}$ and the $a_{0}$ co-efficient's for the series and solving the integral.
The period of $sin^2(t)$ is $\pi$ so I use the interval, $[-\pi/2,\pi/2]$
$$a_{n} = \int_{-\frac\pi2}^\frac\pi2 \sin^2(t)\cos(\frac{k\pi t}{\pi/2})dt$$
So, using the given expression for $a_{n}$ I reduce it to :
$$ a_{n} = \frac{1}{\pi} \left(\int_{-\frac\pi2}^\frac\pi2 cos(2t) dt - \int_{-\frac\pi2}^\frac\pi2 cos(2t)cos(2kt)dt \right)$$
The first integral = 0,
For the second I find,
$$a_{n} = \frac{-1}{2\pi}\left( \frac{sin(\pi + \pi k)}{(k+1)} + \frac{sin(\pi - \pi k)}{(k-1)} \right)$$ But from here I don't know how I proceed to find the complete Fourier series, can anyone help me to finish the problem or to show any mistakes I've made or if there is a simpler way to do the integration.