You don't need equations to work this out. You can just think about it.
You already know that the area under the sine, from \$0\le\theta\le\pi\$ is 2. This is the sum of all the little tiny rectangular areas defined by, on their vertical side by a finite result of the sine function itself (the shape of which is invariant over a half-cycle), times the infinitesimal on the horizontal that is simply \$\text{d}\theta\$.
The integral is just the sum of an infinite number of infinitely thin rectangles stacked side to side.
Now, you can rescale the infinitesimal itself in any way you like. So, instead of an infinite number of \$\text{d}\theta\$ wide rectangles stacked side by side to span between \$0\$ and \$\pi\$, you could just as well instead use an infinite number of \$\text{d} t\$ wide rectangles stacked side by side to span between \$0\$ and \$\frac1{f}\$. The height of each is the same (defined by the sine.) It's just the width of each slice that varies. But since the relative width between them is some constant you create, \$\frac{\text{d}\theta}{\text{d}t}\$ (or \$\frac{\text{d}t}{\text{d}\theta}\$ if you prefer), then the area obviously scales, accordingly.
So it is no surprise to anyone that the sum across two half-cycles of a "squeezed" sine wave integral would be equal to the sum across one half-cycle of an "unsqueezed" sine wave integral. You could just as well take the two "squeezed" half-cycles, unpacking each tiny rectangle that are \$\text{d}t^{\,'}\$ wide, and then restack them so that two \$\text{d}t^{\,'}\$ wide rectangles (one from each of the two half-cycles) are combined to create a single rectangle that is \$\text{d}t\$ wide (where \$\text{d}t=2\:\text{d}t^{\,'}\$ or \$\frac{\text{d}\,t}{\text{d}t^{\,'}}=2\$) and use that new rectangle to add them back up in order to calculate the area for the resulting "unsqueezed" half-cycle.
The result isn't even hard to visualize.