`int2sinxcosxdx` is equal to

Options

A. `cos2x+c`
B. `sin2x+c`
C. `cos^2x+c`
D. `sin^2x+c`

Explanation

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

### Step 1: Use the Double Angle Identity
We know from trigonometric identities that:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite the integral:
\[ \int 2 \sin x \cos x \, dx = \int \sin 2x \, dx \]

### Step 2: Integrate the Function
Now we can integrate \( \sin 2x \):
\[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C \]
Here, \( C \) is the constant of integration.

### Step 3: Simplify the Result
We can express the result in a different form using the cosine double angle identity:
\[ \cos 2x = 1 - 2 \sin^2 x \]
Substituting this into our integral gives:
\[ -\frac{1}{2} \cos 2x = -\frac{1}{2}(1 - 2 \sin^2 x) = -\frac{1}{2} + \sin^2 x \]
Thus, we can rewrite our integral as:
\[ -\frac{1}{2} + \sin^2 x + C \]

### Step 4: Combine Constants
We can combine the constants:
Let \( k = C - \frac{1}{2} \), then:
\[ \int 2 \sin x \cos x \, dx = \sin^2 x + k \]

### Final Result
Thus, the final answer is:
\[ \int 2 \sin x \cos x \, dx = \sin^2 x + C \]
where \( C \) is an arbitrary constant.