The primitive of the function `f(x)=(2x+1)|sin x|`, when `pi lt x lt 2 pi` is

Options

A. `-(2x+1)cos x + 2 sin x +C`
B. `(2x+1)cos x - 2 sin x +C`
C. `(x^(2)+x)cos x +C`
D. None of the above

Answer

To find the primitive (indefinite integral) of the function \( f(x) = (2x + 1)|\sin x| \) in the interval \( \pi < x < 2\pi \), we will follow these steps:

### Step 1: Analyze the function
In the interval \( \pi < x < 2\pi \), the sine function is negative. Therefore, we can express the absolute value of sine as:
\[ |\sin x| = -\sin x \]
Thus, we can rewrite the function \( f(x) \) as:
\[ f(x) = (2x + 1)(-\sin x) = - (2x + 1) \sin x \]

### Step 2: Set up the integral
Now we need to find the integral of \( f(x) \):
\[ \int f(x) \, dx = \int - (2x + 1) \sin x \, dx \]

### Step 3: Use integration by parts
We will use integration by parts, which states:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = 2x + 1 \) (then \( du = 2 \, dx \))
- \( dv = -\sin x \, dx \) (then \( v = \cos x \))

### Step 4: Apply integration by parts
Now we apply the integration by parts formula:
\[ \int - (2x + 1) \sin x \, dx = (2x + 1) \cos x - \int \cos x \cdot 2 \, dx \]
Calculating the integral on the right:
\[ \int \cos x \cdot 2 \, dx = 2 \sin x \]
Thus, we have:
\[ \int - (2x + 1) \sin x \, dx = (2x + 1) \cos x - 2 \sin x + C \]

### Final Result
The primitive of the function \( f(x) \) in the interval \( \pi < x < 2\pi \) is:
\[ \int f(x) \, dx = (2x + 1) \cos x - 2 \sin x + C \]