`int (sin2x)/(sinx )dx = ?`

Options

A. `2 sin x + C`
B. `1/2 sin x + C`
C. `2 cos x + C`
D. `1/2 cos x + C`

Answer

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

### Step 1: Use the double angle identity for sine
We know that:
\[ \sin 2x = 2 \sin x \cos x \]
Substituting this into the integral gives:
\[ \int \frac{\sin 2x}{\sin x} \, dx = \int \frac{2 \sin x \cos x}{\sin x} \, dx \]

### Step 2: Simplify the expression
The \( \sin x \) in the numerator and denominator cancels out:
\[ \int \frac{2 \sin x \cos x}{\sin x} \, dx = \int 2 \cos x \, dx \]

### Step 3: Factor out the constant
We can factor out the constant 2 from the integral:
\[ \int 2 \cos x \, dx = 2 \int \cos x \, dx \]

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

### Step 5: Combine the results
Now, substituting back into our expression gives:
\[ 2 \int \cos x \, dx = 2(\sin x + C) = 2 \sin x + 2C \]
We can simply denote \( 2C \) as \( C \) (since \( C \) is an arbitrary constant):
\[ 2 \sin x + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{\sin 2x}{\sin x} \, dx = 2 \sin x + C \]