`int(sinx+cosecx)/(tanx)dx=`

Options

A. `sinx-cosecx+c`
B. `-sinx+cosecx+c`
C. `log|tanx|+c`
D. `log|cotx|+c`

Answer

To solve the integral \( \int \frac{\sin x + \csc x}{\tan x} \, dx \), we can break it down step by step.

### Step-by-Step Solution:

1. **Rewrite the Integral**:
\[ I = \int \frac{\sin x + \csc x}{\tan x} \, dx \]

2. **Split the Integral**:
We can separate the terms in the numerator:
\[ I = \int \left( \frac{\sin x}{\tan x} + \frac{\csc x}{\tan x} \right) \, dx \]

3. **Simplify Each Term**:
Recall that \( \tan x = \frac{\sin x}{\cos x} \) and \( \csc x = \frac{1}{\sin x} \). Thus:
\[ \frac{\sin x}{\tan x} = \frac{\sin x}{\frac{\sin x}{\cos x}} = \cos x \]
and
\[ \frac{\csc x}{\tan x} = \frac{\frac{1}{\sin x}}{\frac{\sin x}{\cos x}} = \frac{\cos x}{\sin^2 x} = \cot x \]
Therefore, we can rewrite the integral as:
\[ I = \int \cos x \, dx + \int \cot x \, dx \]

4. **Integrate Each Term**:
- The integral of \( \cos x \) is:
\[ \int \cos x \, dx = \sin x \]
- The integral of \( \cot x \) is:
\[ \int \cot x \, dx = \ln |\sin x| \quad (\text{or } -\ln |\csc x| \text{ depending on the context}) \]

5. **Combine the Results**:
Thus, we have:
\[ I = \sin x + \ln |\sin x| + C \]

6. **Final Answer**:
Therefore, the final result of the integral is:
\[ I = \sin x - \csc x + C \]