`int sin mx dx` is:

Options

A. m cos mx+c
B. `-m cos mx+c`
C. `-1/m cos mx +c`
D. `1/m cos mx+c`

Answer

To solve the integral \( \int \sin(mx) \, dx \), we can follow these steps:

### Step 1: Set up the integral
Let:
\[ I = \int \sin(mx) \, dx \]

### Step 2: Use substitution
We will use the substitution \( u = mx \). Then, we differentiate \( u \) with respect to \( x \):
\[ du = m \, dx \quad \Rightarrow \quad dx = \frac{du}{m} \]

### Step 3: Substitute in the integral
Now, substitute \( u \) and \( dx \) into the integral:
\[ I = \int \sin(u) \cdot \frac{du}{m} \]
This simplifies to:
\[ I = \frac{1}{m} \int \sin(u) \, du \]

### Step 4: Integrate \( \sin(u) \)
The integral of \( \sin(u) \) is:
\[ \int \sin(u) \, du = -\cos(u) + C \]
where \( C \) is the constant of integration.

### Step 5: Substitute back for \( u \)
Now, substitute back \( u = mx \):
\[ I = \frac{1}{m} \left( -\cos(mx) + C \right) \]
This simplifies to:
\[ I = -\frac{1}{m} \cos(mx) + C \]

### Final Result
Thus, the integral \( \int \sin(mx) \, dx \) is:
\[ \int \sin(mx) \, dx = -\frac{1}{m} \cos(mx) + C \]

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