The integral `intsqrt(cotx)e^(sqrt(sinx))sqrt(cosx)dx` equals

Options

A. `(sqrt(tanx)e^(sqrt(sinx)))/(sqrt(cosx))+C`
B. `2e^(sqrt(sinx))+C`
C. `-(1)/(2)e^(sqrtsinx)+C`
D. `(sqrt(cotx)e^(sqrt(sinx)))/(2sqrt(cosx))+C`

Explanation

To solve the integral \( I = \int \sqrt{\cot x} e^{\sqrt{\sin x}} \sqrt{\cos x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start by rewriting \(\sqrt{\cot x}\) in terms of sine and cosine:
\[ \sqrt{\cot x} = \sqrt{\frac{\cos x}{\sin x}} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}. \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} e^{\sqrt{\sin x}} \sqrt{\cos x} \, dx. \]
This simplifies to:
\[ I = \int \frac{\cos x}{\sqrt{\sin x}} e^{\sqrt{\sin x}} \, dx. \]

### Step 2: Substitution
Now, we will use the substitution \( t = \sqrt{\sin x} \). Then, we differentiate:
\[ t^2 = \sin x \implies 2t \, dt = \cos x \, dx \implies dx = \frac{2t}{\cos x} \, dt. \]
We also need to express \(\cos x\) in terms of \(t\):
\[ \cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - t^4}. \]
Substituting these into the integral gives:
\[ I = \int \frac{\sqrt{1 - t^4}}{t} e^t \cdot \frac{2t}{\sqrt{1 - t^4}} \, dt. \]
This simplifies to:
\[ I = 2 \int e^t \, dt. \]

### Step 3: Integrate
Now we can integrate:
\[ I = 2 e^t + C. \]

### Step 4: Substitute back
Finally, we substitute back \( t = \sqrt{\sin x} \):
\[ I = 2 e^{\sqrt{\sin x}} + C. \]

### Final Answer
Thus, the integral evaluates to:
\[ I = 2 e^{\sqrt{\sin x}} + C. \]