If `y=sin(2x+3)` then `intydx` will be:

Options

A. `(cos(2x+3))/(2)`
B. `-(cos(2x+3))/(2)+C`
C. `-cos(2x+3)`
D. `-2cos(2x+3)`

Explanation

To solve the integral of \( y = \sin(2x + 3) \), we need to find \( \int y \, dx \).

### Step-by-step Solution:

1. **Identify the integral**:
We start with the integral:
\[ \int \sin(2x + 3) \, dx \]

2. **Substitution**:
To simplify the integral, we can use substitution. Let:
\[ t = 2x + 3 \]
Then, we differentiate \( t \) with respect to \( x \):
\[ dt = 2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{2} \]

3. **Rewrite the integral**:
Substitute \( t \) and \( dx \) into the integral:
\[ \int \sin(2x + 3) \, dx = \int \sin(t) \cdot \frac{dt}{2} \]
This simplifies to:
\[ \frac{1}{2} \int \sin(t) \, dt \]

4. **Integrate**:
The integral of \( \sin(t) \) is:
\[ \int \sin(t) \, dt = -\cos(t) \]
Therefore, we have:
\[ \frac{1}{2} \int \sin(t) \, dt = \frac{1}{2} (-\cos(t)) = -\frac{1}{2} \cos(t) \]

5. **Back-substitute**:
Now, we substitute back \( t = 2x + 3 \):
\[ -\frac{1}{2} \cos(t) = -\frac{1}{2} \cos(2x + 3) \]

6. **Add the constant of integration**:
Finally, we add the constant of integration \( C \):
\[ \int \sin(2x + 3) \, dx = -\frac{1}{2} \cos(2x + 3) + C \]

### Final Answer:
\[ \int \sin(2x + 3) \, dx = -\frac{1}{2} \cos(2x + 3) + C \]