sinusoidal signal, computing its power
Start from:

\$\frac{A^2ω_0}{2π}\int_0^{\frac{2π}{ω_0}}\frac{1}2[1+cos(2ω_0t+2θ)]dt\$

Extract the 1/2 out of the integral, you get 

\$ \frac{A^2ω_0}{2π} \frac12 \frac{2π}{ω_0} 
+ \frac{A^2ω_0}{2π} \int_0^{\frac{2π}{ω_0}}cos(2ω_0t+2θ) \,dt\$

The one on the left is \$ \frac{A^2}{2} \$ and the one on the right is 0 since the cos is integrated over a full period. The mistake is on your third line, because this:

\$ sin(2ω_0t+2θ)\vert_0^\frac{2π}{ω_0}\$

...expands to:

\$ sin(2ω_0\frac{2π}{ω_0}+2θ) - sin(2θ) = sin(4π+2θ) - sin(2θ)\$

(you forgot one of the two) and \$ sin( whatever + 4π ) \$ is equal to \$ sin( whatever ) \$ ...

so this part is zero and you only keep \$ \frac{A^2}{2} \$.