The primitive of the function f (x) `=(2x+1)|cosx|`, when `(pi)/(2)ltxltpi` is given by

Options

A. `cosx+x sin x`
B. `-cosx-xsinx`
C. `xsinx-cos x`
D. none of these

Answer

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

### Step 1: Determine the expression for \( f(x) \)
In the interval \( \frac{\pi}{2} < x < \pi \), the cosine function is negative. Therefore, the absolute value of cosine can be expressed as:
\[ |\cos x| = -\cos x \]
Thus, we can rewrite \( f(x) \) as:
\[ f(x) = (2x + 1)(-\cos x) = -(2x + 1)\cos x \]

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

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

Now we can apply integration by parts:
\[ \int -(2x + 1) \cos x \, dx = -\left[(2x + 1) \sin x - \int \sin x \cdot 2 \, dx\right] \]

### Step 4: Compute the integral
Now we compute the integral:
\[ = -\left[(2x + 1) \sin x - 2 \int \sin x \, dx\right] \]
The integral of \( \sin x \) is:
\[ \int \sin x \, dx = -\cos x \]
So we have:
\[ = -\left[(2x + 1) \sin x - 2(-\cos x)\right] \]
\[ = -\left[(2x + 1) \sin x + 2\cos x\right] \]
\[ = -(2x + 1) \sin x - 2\cos x \]

### Step 5: Add the constant of integration
Finally, we include the constant of integration \( C \):
\[ \int f(x) \, dx = -(2x + 1) \sin x - 2\cos x + C \]

### Final Answer
Thus, the primitive of the function \( f(x) = (2x + 1) |\cos x| \) in the interval \( \frac{\pi}{2} < x < \pi \) is:
\[ -(2x + 1) \sin x - 2\cos x + C \]