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 problem where \( y = \sin(2x + 3) \) and we need to find \( \int y \, dx \), we can follow these steps:

### Step 1: Set up the integral
We start with the integral we need to solve:
\[ \int y \, dx = \int \sin(2x + 3) \, dx \]

### Step 2: Use substitution
To simplify the integral, we can use a substitution. Let:
\[ t = 2x + 3 \]
Now, differentiate both sides with respect to \( x \):
\[ dt = 2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{2} \]

### Step 3: Substitute in the integral
Now we can substitute \( t \) and \( dx \) into the integral:
\[ \int \sin(2x + 3) \, dx = \int \sin(t) \cdot \frac{dt}{2} \]

### Step 4: Factor out the constant
We can factor out the constant \( \frac{1}{2} \):
\[ \int \sin(t) \cdot \frac{dt}{2} = \frac{1}{2} \int \sin(t) \, dt \]

### Step 5: Integrate
Now, we can integrate \( \sin(t) \):
\[ \int \sin(t) \, dt = -\cos(t) \]
Thus, we have:
\[ \frac{1}{2} \int \sin(t) \, dt = \frac{1}{2} (-\cos(t)) = -\frac{1}{2} \cos(t) \]

### Step 6: Substitute back for \( t \)
Now we substitute back \( t = 2x + 3 \):
\[ -\frac{1}{2} \cos(t) = -\frac{1}{2} \cos(2x + 3) \]

### Step 7: Add the constant of integration
Finally, we add the constant of integration \( C \):
\[ \int y \, dx = -\frac{1}{2} \cos(2x + 3) + C \]

### Final Answer
Thus, the final answer is:
\[ \int y \, dx = -\frac{1}{2} \cos(2x + 3) + C \]

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