Question:

Step 1: Define the integral and simplify the integrand. Let the given integral be $I$. We are asked to evaluate: $$ I = \int \sqrt{\cot x} \, e^{\sqrt{\sin x}} \sqrt{\cos x} \, dx $$ We use the trigonometric identity $\cot x = \frac{\cos x}{\sin x}$, so $\sqrt{\cot x} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}$. Substitute this into the integral: $$ I = \int \left( \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \right) e^{\sqrt{\sin x}} \sqrt{\cos x} \, dx $$ Multiply the terms involving $\cos x$: $$ I = \int \frac{\cos x}{\sqrt{\sin x}} e^{\sqrt{\sin x}} \, dx $$ Step 2: Apply a suitable substitution. To simplify the integral, let's substitute $u$ for the expression inside the exponential and the square root. Let: $$ u = \sqrt{\sin x} $$ Step 3: Differentiate the substitution and find $du$. Differentiate $u = \sqrt{\sin x}$ with respect to $x$: $$ \frac{du}{dx} = \frac{d}{dx} (\sin x)^{1/2} $$ Using the chain rule, $\frac{d}{dx} f(g(x)) = f'(g(x))g'(x)$: $$ \frac{du}{dx} = \frac{1}{2} (\sin x)^{-1/2} \cdot (\cos x) $$ $$ \frac{du}{dx} = \frac{\cos x}{2\sqrt{\sin x}} $$ Rearrange to express $dx$ or a part of the integrand in terms of $du$: $$ du = \frac{\cos x}{2\sqrt{\sin x}} \, dx $$ From this, we can see that $\frac{\cos x}{\sqrt{\sin x}} \, dx = 2 \, du$. Step 4: Substitute into the integral and evaluate. Substitute $u = \sqrt{\sin x}$ and $\frac{\cos x}{\sqrt{\sin x}} \, dx = 2 \, du$ into the integral $I$: $$ I = \int e^u (2 \, du) $$ $$ I = 2 \int e^u \, du $$ Now, integrate with respect to $u$: $$ I = 2e^u + C $$ where $C$ is the constant of integration. Step 5: Substitute back the original variable and state the final answer. Substitute $u = \sqrt{\sin x}$ back into the expression for $I$: $$ I = 2e^{\sqrt{\sin x}} + C $$ This matches Option 2. The final answer is $\boxed{\text{2e^\sqrt{sin} x + C}}$.