1. **State the problem:** Evaluate the integral $$\int \frac{4}{\sqrt{16 - 4 \sin^2 \theta}} \, d\theta.$$\n\n2. **Rewrite the integral:** Factor the expression under the square root:\n$$\sqrt{16 - 4 \sin^2 \theta} = \sqrt{4(4 - \sin^2 \theta)} = 2 \sqrt{4 - \sin^2 \theta}.$$\nSo the integral becomes:\n$$\int \frac{4}{2 \sqrt{4 - \sin^2 \theta}} \, d\theta = \int \frac{2}{\sqrt{4 - \sin^2 \theta}} \, d\theta.$$\n\n3. **Use substitution:** Given $x = 2 \sin \theta$, then $\sin \theta = \frac{x}{2}$. Also, $d\theta = \frac{dx}{2 \cos \theta}$. Since $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{x^2}{4}} = \frac{\sqrt{4 - x^2}}{2}$, we have:\n$$d\theta = \frac{dx}{2 \cdot \frac{\sqrt{4 - x^2}}{2}} = \frac{dx}{\sqrt{4 - x^2}}.$$\n\n4. **Rewrite the integral in terms of $x$:**\n$$\int \frac{2}{\sqrt{4 - \sin^2 \theta}} \, d\theta = \int \frac{2}{\sqrt{4 - \frac{x^2}{4}}} \cdot \frac{dx}{\sqrt{4 - x^2}}.$$\nNote that:\n$$\sqrt{4 - \frac{x^2}{4}} = \sqrt{\frac{16 - x^2}{4}} = \frac{\sqrt{16 - x^2}}{2}.$$\nSo the integrand becomes:\n$$\frac{2}{\frac{\sqrt{16 - x^2}}{2}} \cdot \frac{1}{\sqrt{4 - x^2}} = \frac{4}{\sqrt{16 - x^2}} \cdot \frac{1}{\sqrt{4 - x^2}}.$$\n\n5. **Simplify the integral:**\n$$\int \frac{4}{\sqrt{16 - x^2} \sqrt{4 - x^2}} \, dx.$$\nThis integral is complicated, so instead, let's return to the original integral and use a trigonometric identity.\n\n6. **Alternative approach:** Use the identity $\sin^2 \theta = 1 - \cos^2 \theta$:\n$$16 - 4 \sin^2 \theta = 16 - 4 (1 - \cos^2 \theta) = 16 - 4 + 4 \cos^2 \theta = 12 + 4 \cos^2 \theta.$$\nSo the integral becomes:\n$$\int \frac{4}{\sqrt{12 + 4 \cos^2 \theta}} \, d\theta = \int \frac{4}{2 \sqrt{3 + \cos^2 \theta}} \, d\theta = \int \frac{2}{\sqrt{3 + \cos^2 \theta}} \, d\theta.$$\n\n7. **Use substitution $t = \sin \theta$:** Then $d\theta = \frac{dt}{\cos \theta} = \frac{dt}{\sqrt{1 - t^2}}$. Also, $\cos^2 \theta = 1 - t^2$. So the integral becomes:\n$$\int \frac{2}{\sqrt{3 + 1 - t^2}} \cdot \frac{dt}{\sqrt{1 - t^2}} = \int \frac{2}{\sqrt{4 - t^2}} \cdot \frac{dt}{\sqrt{1 - t^2}} = 2 \int \frac{dt}{\sqrt{(4 - t^2)(1 - t^2)}}.$$\n\n8. **This integral is an elliptic integral, which generally cannot be expressed in elementary functions.**\n\n**Final answer:** The integral simplifies to an elliptic integral form:\n$$\int \frac{4}{\sqrt{16 - 4 \sin^2 \theta}} \, d\theta = 2 \int \frac{dt}{\sqrt{(4 - t^2)(1 - t^2)}} + C,$$\nwhere $t = \sin \theta$.