Evaluate: `intsec(2x+3)tan(2x+3)dx`

Explanation

To evaluate the integral \( \int \sec(2x + 3) \tan(2x + 3) \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \sec(2x + 3) \tan(2x + 3) \, dx \]
We know that:
\[ \sec(x) = \frac{1}{\cos(x)} \quad \text{and} \quad \tan(x) = \frac{\sin(x)}{\cos(x)} \]
Thus, we can rewrite the integral as:
\[ \int \frac{\sin(2x + 3)}{\cos(2x + 3)} \cdot \frac{1}{\cos(2x + 3)} \, dx = \int \frac{\sin(2x + 3)}{\cos^2(2x + 3)} \, dx \]

### Step 2: Use Substitution
Let’s use substitution to simplify the integral. We set:
\[ t = \cos(2x + 3) \]
Then, we differentiate both sides:
\[ \frac{dt}{dx} = -2 \sin(2x + 3) \quad \Rightarrow \quad dt = -2 \sin(2x + 3) \, dx \quad \Rightarrow \quad dx = \frac{dt}{-2 \sin(2x + 3)} \]

### Step 3: Substitute in the Integral
Substituting \( t \) and \( dx \) into the integral, we have:
\[ \int \frac{\sin(2x + 3)}{t^2} \cdot \frac{dt}{-2 \sin(2x + 3)} = \int \frac{1}{-2t^2} \, dt \]

### Step 4: Integrate
Now we can integrate:
\[ \int \frac{1}{-2t^2} \, dt = -\frac{1}{2} \int t^{-2} \, dt = -\frac{1}{2} \left( -\frac{1}{t} \right) = \frac{1}{2t} + C \]

### Step 5: Substitute Back
Now, we substitute back \( t = \cos(2x + 3) \):
\[ \frac{1}{2 \cos(2x + 3)} + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \sec(2x + 3) \tan(2x + 3) \, dx = \frac{1}{2 \cos(2x + 3)} + C \]