I'm pretty sure this is correct since I'm able to justify each of the steps but I'm just wondering if there's a more efficient way that isn't just skipping steps? Because I had to do 2 substitutions, one of which converted sine into cosine.
\begin{align*} \text{Length }= & \ 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{2 + 2\sin\left(\theta\right)} d\theta \\ = & \ 2 \sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1 + \sin\left(\theta\right)} d\theta\\ = & \ - 2\sqrt{2} \int_{\pi}^{0} \sqrt{1+ \sin\left(\frac{\pi}{2}-x\right) } dx \\ = & \ - 2\sqrt{2} \int_{\pi}^{0} \sqrt{1+ \cos\left(x\right) } dx \\ = & \ - 2\sqrt{2} \int_{\pi}^{0} \sqrt{2 \cos^2\left(\frac{x}{2}\right)} dx \\ = & \ - 4 \int_{\pi}^{0} \left|\cos\left(\frac{x}{2}\right)\right| dx \\ = & \ 4 \int_{0}^{\pi} \cos\left(\frac{x}{2}\right) dx \\ = & \ 4 \cdot 2 \int_{0}^{\frac{\pi}{2}} \cos\left(u\right) du \\ = & \ 8 \left[\sin\left(u\right)\right]_{u=0}^{u=\frac{\pi}{2}} \\ = & \ 8 \left[\sin\left(\frac{\pi}{2}\right) - \sin\left(0\right) \right] \\ = & \ 8 \\ \end{align*}
edit: admittedly, I do know of other answers but I have no idea how one would have come across such trigonometric identities, or reason out how such an identity would be beneficial.
Instead
\begin{align*} \text{L }= & \ 2 \sqrt2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1 +\sin\theta}\> d\theta = \ 2 \sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\left(\sin\frac{\theta}2+ \cos\frac{\theta}2 \right)^2} \>d\theta\\ = &\ 2\sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\sin\frac{\theta}2+ \cos\frac{\theta}2 \right) d\theta = 8 \\ \end{align*}