Basic question about voltage
First thing to note is that the step function is defined as

$$
  u(t) = \left.\begin{cases} 0 &amp; t&lt;0\\ 1 &amp; t&gt;0\\ \end{cases}\right\}
$$

so 

$$
  V_c = \frac{1}{C}\int_{-\infty}^{t_f} i(t) dt
$$

$$
  V_c = \frac{1}{C}\int_{-\infty}^{0} 0.4\sin (2t - \frac{\pi}{4} )u(t) dt + \frac{1}{C}\int_0^{t_f} 0.4\sin (2t - \frac{\pi}{4} )u(t) dt
$$

$$
  V_c = \frac{0.4}{C}\int_0^{t_f} \sin (2t - \frac{\pi}{4} ) dt
$$

So all the step function does is set all values of t that are less than zero to zero.

You then continue to compute the integral in the normal way, so:

$$
   V_c = \frac{0.4}{C} \left[ \frac{1}{2} \cos(2t - \frac{\pi}{4})\right]_0^{t_f}
$$