`int (cos x)/( sin x+sqrt(sin x)) dx`

Answer

To solve the integral \(\int \frac{\cos x}{\sin x + \sqrt{\sin x}} \, dx\), we can follow these steps:

### Step 1: Substitution
Let \( t = \sqrt{\sin x} \). Then, we have:
\[ \sin x = t^2 \]
Now, we need to find \( dx \) in terms of \( dt \).

### Step 2: Differentiate \( t \)
Differentiating both sides with respect to \( x \):
\[ \frac{dt}{dx} = \frac{1}{2\sqrt{\sin x}} \cdot \cos x = \frac{1}{2t} \cos x \]
Thus, we can express \( dx \) as:
\[ dx = \frac{2t}{\cos x} \, dt \]

### Step 3: Substitute in the Integral
Now substitute \( \sin x \) and \( dx \) into the integral:
\[ \int \frac{\cos x}{t^2 + t} \cdot \frac{2t}{\cos x} \, dt \]
The \(\cos x\) terms cancel out:
\[ = \int \frac{2t}{t^2 + t} \, dt \]

### Step 4: Simplify the Integral
We can simplify the integrand:
\[ = \int \frac{2t}{t(t + 1)} \, dt = \int \frac{2}{t + 1} \, dt \]

### Step 5: Integrate
Now we can integrate:
\[ = 2 \int \frac{1}{t + 1} \, dt = 2 \ln |t + 1| + C \]

### Step 6: Substitute Back
Now substitute back \( t = \sqrt{\sin x} \):
\[ = 2 \ln |1 + \sqrt{\sin x}| + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int \frac{\cos x}{\sin x + \sqrt{\sin x}} \, dx = 2 \ln |1 + \sqrt{\sin x}| + C \]

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