I have the answer to this problem. My question is with the function $u(t)$. $u(t)$ is:
$$u(t) = 2\cos(t)+2\sin(t-\pi/2)*1(t-\pi/2)$$
Why is there a $1(t-\pi/2)$ multiplying the $2\sin(t-\pi/2)$? Wouldn't $u(t) = 2\cos(t)+2\sin(t-\pi/2)$ suffice for $ u(t)$? $1(t-\pi/2)$ is a step function right? What's that got to do with a $\sin$? I'm very confused about this. Any help would be appreciated.
Note that $\cos(t) + \sin(t-\pi/2) = 0$ and $\sin$ is needed to take effect after the point $\pi/2$ to obtain the given graphic. Hence it is multiplied by $1(t-\pi/2)$ which is the unit step function as you correctly guessed.
In other words you need to obtain
$$u(t)=\begin{cases} \cos(t) & 0<t<\pi/2 \\ 0 & \pi/2 \leq t \end{cases}$$
and remember the definition of step function.