Question:

Step 1: Rewrite the integrand using trigonometric identities. The given integral is $I = \int \sqrt{\cot x} e^{\sqrt{\sin x}} \sqrt{\cos x} dx$. We know that $\sqrt{\cot x} = \sqrt{\frac{\cos x}{\sin x}} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}$. Substitute this into the integral: $$I = \int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} e^{\sqrt{\sin x}} \sqrt{\cos x} dx$$ Combine the terms involving $\sqrt{\cos x}$: $$I = \int \frac{\cos x}{\sqrt{\sin x}} e^{\sqrt{\sin x}} dx$$ Step 2: Apply a suitable substitution to simplify the integral. Let $u = \sqrt{\sin x}$. This substitution is chosen because its derivative involves terms present in the rest of the integrand. Step 3: Calculate the differential $du$ in terms of $dx$ and rearrange. Differentiate $u = \sqrt{\sin x}$ with respect to $x$: $$\frac{du}{dx} = \frac{d}{dx} (\sin x)^{1/2} = \frac{1}{2} (\sin x)^{-1/2} \cdot \cos x$$ $$\frac{du}{dx} = \frac{\cos x}{2\sqrt{\sin x}}$$ Now, express $dx$ or a part of the integrand in terms of $du$: $$du = \frac{\cos x}{2\sqrt{\sin x}} dx$$ Rearranging, we get: $$\frac{\cos x}{\sqrt{\sin x}} dx = 2 du$$ Step 4: Substitute $u$ and $du$ into the integral and evaluate. Substitute $u = \sqrt{\sin x}$ and $\frac{\cos x}{\sqrt{\sin x}} dx = 2 du$ into the integral $I$: $$I = \int e^u (2 du)$$ Factor out the constant and integrate: $$I = 2 \int e^u du$$ $$I = 2e^u + C$$ where $C$ is the constant of integration. Step 5: Substitute back the original variable and state the final answer. Replace $u$ with $\sqrt{\sin x}$: $$I = 2e^{\sqrt{\sin x}} + C$$ Comparing this result with the given options, we find that it matches Option 2. The final answer is $\boxed{2e^{\sqrt{\sin x}} + C}$.