Why we don't integrate intital velocity in body cast equation?

On this site I've found a formula for calculating the $x, y$ coordinates for a body throwed by an angle to a horizon.

It looks like this:
$$x(t) = V_0 t \cos(\alpha);$$ $$y(t) = V_0 t \sin(\alpha) - gt^2/2;$$

It's clear to me that term $(g t^2)/2$ was received by the integration of $gt \, dt$.
But, I don't understand, why didn't we integrate $V_0 t \sin(\alpha)$ part of the equation?
Why do we integrate only $gt$ term?

• Everything were integrated. That's why there is a $t$ in $v_0 t\sin{a}$. – Physicist137 Nov 19 '14 at 23:51
• @Physicist137, so, the original equation was then x(t)=V0*sin(a), is that's right? – PaulD Nov 20 '14 at 0:06
• Well, no. You must integrate velocity to get position. Hence, original equation was: $v_y(t) = v_0\sin a - gt$ and $v_x = v_0\cos a$. – Physicist137 Nov 20 '14 at 0:14
• @Physicist137, now I understand, thanks. – PaulD Nov 20 '14 at 0:22
• $\int V_0 sin(\alpha) dt = V_0 sin(\alpha) t$ – Jold Nov 20 '14 at 0:36