Evaluate : `int " cosec"x dx`

Answer

To evaluate the integral \( \int \csc x \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \csc x \, dx \]

### Step 2: Multiply and Divide by \( \csc x - \cot x \)
To simplify the integral, we multiply and divide by \( \csc x - \cot x \):
\[ I = \int \csc x \cdot \frac{\csc x - \cot x}{\csc x - \cot x} \, dx \]
This gives us:
\[ I = \int \frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x} \, dx \]

### Step 3: Substitution
Let:
\[ t = \csc x - \cot x \]
Now, we need to find \( dt \):
\[ dt = (-\csc x \cot x + \csc^2 x) \, dx \]
This simplifies to:
\[ dt = \csc^2 x - \csc x \cot x \, dx \]

### Step 4: Solve for \( dx \)
From the expression for \( dt \), we can express \( dx \):
\[ dx = \frac{dt}{\csc^2 x - \csc x \cot x} \]

### Step 5: Substitute Back into the Integral
Now substituting back into the integral:
\[ I = \int \frac{1}{t} \, dt \]

### Step 6: Integrate
The integral of \( \frac{1}{t} \) is:
\[ I = \ln |t| + C \]
Substituting back for \( t \):
\[ I = \ln |\csc x - \cot x| + C \]

### Final Result
Thus, the final result for the integral \( \int \csc x \, dx \) is:
\[ \int \csc x \, dx = \ln |\csc x - \cot x| + C \]

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