`int [cosec (log x)] [1-cot (log x)] dx`

Answer

To solve the integral \( \int \csc(\log x) (1 - \cot(\log x)) \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \csc(\log x) (1 - \cot(\log x)) \, dx \]
We can distribute the terms inside the integral:
\[ \int \csc(\log x) \, dx - \int \csc(\log x) \cot(\log x) \, dx \]

### Step 2: Substitution
Let \( t = \log x \). Then, the differential \( dx \) can be expressed as:
\[ dx = e^t \, dt \]
Now, substituting \( t \) into the integral gives us:
\[ \int \csc(t) \, e^t \, dt - \int \csc(t) \cot(t) \, e^t \, dt \]

### Step 3: Solve the First Integral
The integral \( \int \csc(t) \, dt \) is a standard integral:
\[ \int \csc(t) \, dt = -\log |\csc(t) + \cot(t)| + C_1 \]

### Step 4: Solve the Second Integral
The integral \( \int \csc(t) \cot(t) \, dt \) is also a standard integral:
\[ \int \csc(t) \cot(t) \, dt = -\csc(t) + C_2 \]

### Step 5: Combine the Results
Now we can combine the results of the two integrals:
\[ -\log |\csc(t) + \cot(t)| + \csc(t) + C \]
Substituting back \( t = \log x \):
\[ -\log |\csc(\log x) + \cot(\log x)| + \csc(\log x) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ -\log |\csc(\log x) + \cot(\log x)| + \csc(\log x) + C \]

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