Equation for the average power of a signal
The integral given is a general formula for the average power of a periodic signal:

$$P_{av}=\frac{1}{T}\int_0^T x^2(t) \mathrm{d}t$$

This equation does not depend on the duty cycle being 50% or the rise and fall times being equal. 

It does depend on you working out a form for \$x(t)\$ that gives the instantaneous power when squared. For example, if you're talking about a voltage signal applied to a resistive load, you'd use \$x(t) = \dfrac{v(t)}{\sqrt{R}}\$ instead of just \$v(t)\$.

Only the second equation, where he reduces the integral to a formula based on the actual signal parameters, depends on those assumptions.