1. **Problem:** Use the Weierstrass substitution method to evaluate the integral $$\int \frac{\sin x}{1 + \cos x} \, dx.$$\n\n2. **Recall the Weierstrass substitution:** Let $$t = \tan\frac{x}{2}$$ which implies:\n$$\sin x = \frac{2t}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2}, \quad dx = \frac{2}{1+t^2} dt.$$\n\n3. **Rewrite the integral in terms of $t$:$$\int \frac{\sin x}{1 + \cos x} dx = \int \frac{\frac{2t}{1+t^2}}{1 + \frac{1-t^2}{1+t^2}} \cdot \frac{2}{1+t^2} dt.$$\n\n4. **Simplify the denominator inside the integral:**\n$$1 + \frac{1-t^2}{1+t^2} = \frac{(1+t^2) + (1 - t^2)}{1+t^2} = \frac{2}{1+t^2}.$$\n\n5. **Substitute back:**\n$$\int \frac{\frac{2t}{1+t^2}}{\frac{2}{1+t^2}} \cdot \frac{2}{1+t^2} dt = \int t \cdot \frac{2}{1+t^2} dt = 2 \int \frac{t}{1+t^2} dt.$$\n\n6. **Integrate:** Use substitution $u = 1 + t^2$, so $du = 2t dt$, hence $t dt = \frac{du}{2}$.\n$$2 \int \frac{t}{1+t^2} dt = 2 \int \frac{t dt}{u} = 2 \int \frac{1}{u} \cdot \frac{du}{2} = \int \frac{1}{u} du = \ln|u| + C = \ln(1 + t^2) + C.$$\n\n7. **Back-substitute $t = \tan \frac{x}{2}$:**\n$$\int \frac{\sin x}{1 + \cos x} dx = \ln\left(1 + \tan^2 \frac{x}{2}\right) + C = \ln\left(\sec^2 \frac{x}{2}\right) + C = 2 \ln\left|\sec \frac{x}{2}\right| + C.$$\n\n**Final answer:** $$\int \frac{\sin x}{1 + \cos x} dx = 2 \ln\left|\sec \frac{x}{2}\right| + C.$$