`int(cos 2x)/(sinx)dx=`

Options

A. `log|tan((x)/(2))|-2cosx+c`
B. `log|tan((x)/(2))|+2cosx+c`
C. `(1)/(2)log|tan((x)/(2))|-2cosx+c`
D. `(1)/(2)log|tan((x)/(2))|+2cosx+c`

Answer

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

### Step 1: Rewrite \(\cos 2x\) in terms of \(\sin x\)
Using the double angle identity for cosine, we have:
\[ \cos 2x = 1 - 2\sin^2 x \]
Thus, we can rewrite the integral:
\[ I = \int \frac{1 - 2\sin^2 x}{\sin x} \, dx \]

### Step 2: Split the integral
Now we can split the integral into two parts:
\[ I = \int \frac{1}{\sin x} \, dx - 2 \int \sin x \, dx \]

### Step 3: Evaluate the first integral
The integral \(\int \frac{1}{\sin x} \, dx\) can be rewritten as:
\[ \int \csc x \, dx \]
The integral of \(\csc x\) is:
\[ \int \csc x \, dx = \ln |\csc x - \cot x| + C \]

### Step 4: Evaluate the second integral
The integral \(\int \sin x \, dx\) is:
\[ \int \sin x \, dx = -\cos x + C \]

### Step 5: Combine the results
Now substituting back into our equation for \(I\):
\[ I = \ln |\csc x - \cot x| - 2(-\cos x) + C \]
This simplifies to:
\[ I = \ln |\csc x - \cot x| + 2\cos x + C \]

### Final Answer
Thus, the final result for the integral is:
\[ I = \ln |\csc x - \cot x| + 2\cos x + C \]

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