Integrate :
`int (sqrt( cos 2x))/(sin x)dx`

Options

A. `log|cot x+ sqrt(cot^(2)x-1)|+sqrt(2)log|cos x +sqrt(cos^(2)x-1//2)|+C`
B. `- log|cot x+ sqrt(cot^(2)x-1)|+sqrt(2)log|cos x +sqrt(cos^(2)x-1//2)|+C`
C. `log|cot x+sqrt(cot^(2)x-1)|+2log|cos x + sqrt(cos^(2)x-1//2)|+C`
D. `- log|cot x+sqrt(cot^(2)x-1)|+2log|cos x + sqrt(cos^(2)x-1//2)|+C`

Answer

To solve the integral \( I = \int \frac{\sqrt{\cos 2x}}{\sin x} \, dx \), we can follow these steps:

### Step 1: Multiply and Divide by \( \cos 2x \)
We start by rewriting the integral:
\[ I = \int \frac{\sqrt{\cos 2x}}{\sin x} \, dx = \int \frac{\cos 2x}{\sin x \sqrt{\cos 2x}} \cdot \frac{1}{\cos 2x} \, dx \]

### Step 2: Substitute \( \cos 2x \)
Using the identity \( \cos 2x = 1 - 2\sin^2 x \), we substitute:
\[ I = \int \frac{1 - 2\sin^2 x}{\sin x \sqrt{1 - 2\sin^2 x}} \, dx \]

### Step 3: Separate the Integral
We can separate the integral into two parts:
\[ I = \int \frac{1}{\sin x \sqrt{\cos 2x}} \, dx - 2 \int \frac{\sin^2 x}{\sin x \sqrt{\cos 2x}} \, dx \]
This simplifies to:
\[ I = \int \frac{1}{\sin x \sqrt{\cos 2x}} \, dx - 2 \int \frac{\sin x}{\sqrt{\cos 2x}} \, dx \]

### Step 4: Solve the First Integral
Let \( I_1 = \int \frac{1}{\sin x \sqrt{\cos 2x}} \, dx \). We can express \( \cos 2x \) as \( \cos^2 x - \sin^2 x \):
\[ I_1 = \int \frac{dx}{\sin x \sqrt{\cos^2 x - \sin^2 x}} \]

### Step 5: Solve the Second Integral
Let \( I_2 = 2 \int \frac{\sin x}{\sqrt{\cos 2x}} \, dx \). We can use the substitution \( u = \cos x \), which gives \( du = -\sin x \, dx \):
\[ I_2 = -2 \int \frac{du}{\sqrt{2u - 1}} \]

### Step 6: Integrate
The integral \( I_2 \) can be solved using the formula for integrals of the form \( \int \frac{du}{\sqrt{a - u}} \):
\[ I_2 = -2 \cdot \sqrt{2} \ln |u + \sqrt{2u - 1}| + C \]

### Step 7: Combine Results
Now, we combine the results of \( I_1 \) and \( I_2 \):
\[ I = I_1 - I_2 = \int \frac{1}{\sin x \sqrt{\cos 2x}} \, dx + 2\sqrt{2} \ln |\cos x + \sqrt{2\cos x - 1}| + C \]

### Final Result
Thus, the final answer is:
\[ I = \int \frac{1}{\sin x \sqrt{\cos 2x}} \, dx + 2\sqrt{2} \ln |\cos x + \sqrt{2\cos x - 1}| + C \]