1. **Problem Statement:** Evaluate the integral $$\int \sqrt{1 + \sin x} \, dx$$. 2. **Formula and Identities Used:** - Use the double-angle identity for sine: $$\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$$. - Use the Pythagorean identity: $$\sin^2 \theta + \cos^2 \theta = 1$$. 3. **Step-by-step Solution:** - Start with the integral: $$\int \sqrt{1 + \sin x} \, dx$$ - Substitute $$\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$$: $$\int \sqrt{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}} \, dx$$ - Rewrite 1 as $$\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}$$: $$\int \sqrt{\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}} \, dx$$ - Recognize the expression inside the square root as a perfect square: $$\int \sqrt{\left(\sin \frac{x}{2} + \cos \frac{x}{2}\right)^2} \, dx$$ - Simplify the square root: $$\int \left|\sin \frac{x}{2} + \cos \frac{x}{2}\right| \, dx$$ - Assuming the expression inside is positive (or considering the antiderivative), write: $$\int \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right) \, dx$$ - Integrate term by term: $$\int \sin \frac{x}{2} \, dx + \int \cos \frac{x}{2} \, dx$$ - Use substitution for each integral: - For $$\int \sin \frac{x}{2} \, dx$$, let $$u = \frac{x}{2}$$, so $$dx = 2 du$$: $$\int \sin u \cdot 2 \, du = 2 \int \sin u \, du = -2 \cos u + C = -2 \cos \frac{x}{2} + C$$ - For $$\int \cos \frac{x}{2} \, dx$$, similarly: $$2 \int \cos u \, du = 2 \sin u + C = 2 \sin \frac{x}{2} + C$$ - Combine results: $$-2 \cos \frac{x}{2} + 2 \sin \frac{x}{2} + C$$ - Factor out 2: $$2 \left(\sin \frac{x}{2} - \cos \frac{x}{2}\right) + C$$ 4. **Final Answer:** $$\int \sqrt{1 + \sin x} \, dx = 2 \left(\sin \frac{x}{2} - \cos \frac{x}{2}\right) + C$$ This solution uses trigonometric identities to simplify the integrand into a perfect square, making the integral straightforward to evaluate.