Fourier components of $\cos(2\pi f_1t)$
HINT:

Going from your last equation, 
$$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)$$
This can be simplified further down by considering the following:

\begin{align}
e^{j2\pi (f_1T-n)} &amp;= e^{j\pi (f_1T-n)}\cdot e^{j\pi (f_1T-n)}\\
1 &amp;=e^{j\pi (f_1T-n)}\cdot e^{-j\pi (f_1T-n)}\\
e^{-j\pi n} &amp;= (-1)^n\\
\end{align}
And that
\begin{align}
\mathrm{sinc}(x) &amp;= \frac{\sin(x)}{x}\\
\sin(x) &amp;= \frac{e^{jx} - e^{-jx}}{2j}
\end{align}