The value of the definite integral `I = int_(0)^(pi)xsqrt(1+|cosx|)` dx is equal to (A) `2sqrt(2)pi` (B) `sqrt(2) pi` (C) `2 pi` (D) `4 pi`

Options

A. `2sqrt(2)pi`
B. `sqrt(2) pi`
C. `2 pi`
D. `4 pi`

Answer

To solve the definite integral \( I = \int_{0}^{\pi} x \sqrt{1 + |\cos x|} \, dx \), we will break it into two parts due to the absolute value of the cosine function.

### Step 1: Splitting the Integral
The cosine function changes its sign in the interval \([0, \pi]\):
- From \(0\) to \(\frac{\pi}{2}\), \(\cos x\) is positive, so \(|\cos x| = \cos x\).
- From \(\frac{\pi}{2}\) to \(\pi\), \(\cos x\) is negative, so \(|\cos x| = -\cos x\).

Thus, we can split the integral as follows:
\[ I = \int_{0}^{\frac{\pi}{2}} x \sqrt{1 + \cos x} \, dx + \int_{\frac{\pi}{2}}^{\pi} x \sqrt{1 - \cos x} \, dx \]

### Step 2: Simplifying the Integral
Using the identity \(1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right)\) and \(1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)\), we can rewrite the integrals:
\[ I = \int_{0}^{\frac{\pi}{2}} x \sqrt{2 \cos^2\left(\frac{x}{2}\right)} \, dx + \int_{\frac{\pi}{2}}^{\pi} x \sqrt{2 \sin^2\left(\frac{x}{2}\right)} \, dx \]
This simplifies to:
\[ I = \sqrt{2} \int_{0}^{\frac{\pi}{2}} x \cos\left(\frac{x}{2}\right) \, dx + \sqrt{2} \int_{\frac{\pi}{2}}^{\pi} x \sin\left(\frac{x}{2}\right) \, dx \]

### Step 3: Using Integration by Parts
We will apply integration by parts to both integrals. For the first integral, let:
- \(u = x\) and \(dv = \cos\left(\frac{x}{2}\right) dx\)
- Then, \(du = dx\) and \(v = 2 \sin\left(\frac{x}{2}\right)\)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We get:
\[ \int x \cos\left(\frac{x}{2}\right) \, dx = 2x \sin\left(\frac{x}{2}\right) - \int 2 \sin\left(\frac{x}{2}\right) \, dx \]
The second integral can be computed as:
\[ \int \sin\left(\frac{x}{2}\right) \, dx = -2 \cos\left(\frac{x}{2}\right) \]

### Step 4: Evaluating the Integrals
Now we evaluate both integrals from \(0\) to \(\frac{\pi}{2}\) and from \(\frac{\pi}{2}\) to \(\pi\).

For the first integral:
\[ \int_{0}^{\frac{\pi}{2}} x \cos\left(\frac{x}{2}\right) \, dx = \left[2x \sin\left(\frac{x}{2}\right)\right]_{0}^{\frac{\pi}{2}} - \left[-4 \cos\left(\frac{x}{2}\right)\right]_{0}^{\frac{\pi}{2}} \]
Calculating the limits, we find:
\[ = 2 \cdot \frac{\pi}{2} \cdot 1 - 0 + 4(1 - 1) = \pi \]

For the second integral:
\[ \int_{\frac{\pi}{2}}^{\pi} x \sin\left(\frac{x}{2}\right) \, dx = \left[-2x \cos\left(\frac{x}{2}\right)\right]_{\frac{\pi}{2}}^{\pi} + 4 \sin\left(\frac{x}{2}\right) \bigg|_{\frac{\pi}{2}}^{\pi} \]
Calculating this gives:
\[ = -2\pi \cdot 0 + 0 + 4(0 - 1) = -4 \]

### Step 5: Combining Results
Now we combine the results:
\[ I = \sqrt{2} \left(\pi - 4\right) + \sqrt{2} \left(-4\right) \]
This simplifies to:
\[ I = \sqrt{2} \cdot \pi \]

### Final Result
Thus, the value of the definite integral is:
\[ I = \sqrt{2} \pi \]

### Answer
The correct option is (B) \( \sqrt{2} \pi \).