Angular velocity: why is integral written with $t$ and with $\tau$? In a physics exercise book I am using, the integral to get the angular velocity $\omega$ from some angular acceleration $\alpha (t)$ is written like this:
$$\omega(t) = \int_0^t \alpha(\tau) d \tau = \alpha_0 \left(t + \frac{2t_1}{\pi} \cos\left(\frac{\pi}{2t_1} t\right) - \frac{2t_1}{\pi} \right) $$
Why is the $\tau$ used here?
 A: The $\tau$ is a dummy variable in the integral.
Anything (other than $t$ used in the limits and $a$ in the integrand) could have been used.
(This is really a math notation question... not a physics one.)
Update: think about slots.
$$\omega( \star)=\int^\star_0 a(\spadesuit) d\spadesuit$$
Instead of
$$\omega( t)=\int^t_0 a(\tau) d\tau$$
one sometimes uses
$$\omega( t)=\int^t_0 a(t') dt'$$
(The $\tau$ resembles the $t'$.)
Note $t$ and $\tau$ have the same units,
but they represent different things.
As used, $\tau$ is used to label how the integrand over the region
is chopped up and integrated. $t$ is used to specified the upper limit of the region.
A: Consider the expression $\sum_{n=1}^5 n^2$.  Think about how you would evaluate that expression.
$$\sum_{n=1}^5 n^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2$$
The summation starts at 1 and ends at 5.  Now try to evaluate this expression:
$$\sum_{n=1}^n n^2$$
Do you see the problem? That's the same reason you can't have something like $\int_0^t f(t) \mathrm dt$.
A: The issue is that $t$ and $\tau$ mean something different in your case.
$t$ seems to be used as a placeholder for a specific value, more precisely the upper value of the interval over which the definite integral is to be computed.
But to integrate you need a variable, not a specific value. So it is a mathematical "trick" to simply use another symbol for the integration variable until the integration process is done.
A: the angular acceleration is
$$\frac{d\omega}{dt}=\alpha(t)$$
from here
$$ \omega(t)=\int \alpha(t) \,dt+c\\
 $$
where the constant c is $~\omega(0)~$
this is equivalence to
$$ \omega(t)=\int_0^t \alpha(\tau) \,d\tau\\
 $$
