`int (sec^(2)x)/(cosec^(2)x) dx`

Answer

To solve the integral \( \int \frac{\sec^2 x}{\csc^2 x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting the integrand using the definitions of secant and cosecant:
\[ \sec^2 x = \frac{1}{\cos^2 x} \quad \text{and} \quad \csc^2 x = \frac{1}{\sin^2 x} \]
Thus, we can rewrite the integral as:
\[ \int \frac{\sec^2 x}{\csc^2 x} \, dx = \int \frac{\frac{1}{\cos^2 x}}{\frac{1}{\sin^2 x}} \, dx = \int \frac{\sin^2 x}{\cos^2 x} \, dx \]

### Step 2: Simplify the integrand
The expression \( \frac{\sin^2 x}{\cos^2 x} \) can be simplified to:
\[ \frac{\sin^2 x}{\cos^2 x} = \tan^2 x \]
So the integral now becomes:
\[ \int \tan^2 x \, dx \]

### Step 3: Use the identity for tangent
We can use the identity \( \tan^2 x = \sec^2 x - 1 \) to rewrite the integral:
\[ \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx \]

### Step 4: Split the integral
Now, we can split the integral into two parts:
\[ \int (\sec^2 x - 1) \, dx = \int \sec^2 x \, dx - \int 1 \, dx \]

### Step 5: Integrate each part
We know that:
\[ \int \sec^2 x \, dx = \tan x + C \]
and
\[ \int 1 \, dx = x \]
Thus, we have:
\[ \int \tan^2 x \, dx = \tan x - x + C \]

### Final Answer
Therefore, the final result of the integral is:
\[ \int \frac{\sec^2 x}{\csc^2 x} \, dx = \tan x - x + C \]

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