`int_(0)^(pi/2)(x)/(1+cos x)dx =`

Options

A. `pi-(log 2)`
B. `(pi)/(4)-(log 2)`
C. `(pi)/(3)-(log 2)`
D. `(pi)/(2)-(log 2)`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \frac{x}{1 + \cos x} \, dx \), we will follow a step-by-step approach.

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{x}{1 + \cos x} \, dx \]

### Step 2: Use the Identity for Cosine
We can use the identity \( \cos x = 1 - 2 \sin^2\left(\frac{x}{2}\right) \) to rewrite \( 1 + \cos x \):
\[ 1 + \cos x = 2 - 2 \sin^2\left(\frac{x}{2}\right) = 2 \left(1 - \sin^2\left(\frac{x}{2}\right)\right) = 2 \cos^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{x}{2 \cos^2\left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} x \sec^2\left(\frac{x}{2}\right) \, dx \]

### Step 3: Integration by Parts
Let \( u = x \) and \( dv = \sec^2\left(\frac{x}{2}\right) \, dx \). Then, we differentiate and integrate:
- \( du = dx \)
- \( v = 2 \tan\left(\frac{x}{2}\right) \) (since \( \int \sec^2(kx) \, dx = \frac{1}{k} \tan(kx) + C \))

Applying integration by parts:
\[ I = \frac{1}{2} \left[ x \cdot 2 \tan\left(\frac{x}{2}\right) \bigg|_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2 \tan\left(\frac{x}{2}\right) \, dx \right] \]

### Step 4: Evaluate the Boundary Terms
Evaluating the boundary term:
\[ x \cdot 2 \tan\left(\frac{x}{2}\right) \bigg|_{0}^{\frac{\pi}{2}} = \left(\frac{\pi}{2} \cdot 2 \tan\left(\frac{\pi}{4}\right)\right) - \left(0 \cdot 2 \tan(0)\right) = \frac{\pi}{2} \cdot 2 \cdot 1 - 0 = \pi \]

### Step 5: Evaluate the Remaining Integral
Now we need to evaluate:
\[ \int_{0}^{\frac{\pi}{2}} 2 \tan\left(\frac{x}{2}\right) \, dx \]
Using the substitution \( u = \frac{x}{2} \), \( dx = 2 \, du \):
\[ = 2 \int_{0}^{\frac{\pi}{4}} \tan(u) \, 2 \, du = 4 \int_{0}^{\frac{\pi}{4}} \tan(u) \, du \]
The integral of \( \tan(u) \) is:
\[ \int \tan(u) \, du = -\log(\cos(u)) + C \]
Thus,
\[ = 4 \left[-\log(\cos(u)) \bigg|_{0}^{\frac{\pi}{4}}\right] = 4 \left[-\log(\cos(\frac{\pi}{4})) + \log(\cos(0))\right] = 4 \left[-\log\left(\frac{1}{\sqrt{2}}\right) + \log(1)\right] = 4 \left[\frac{1}{2} \log(2)\right] = 2 \log(2) \]

### Step 6: Combine Results
Now substituting back into our expression for \( I \):
\[ I = \frac{1}{2} \left[\pi - 2 \log(2)\right] = \frac{\pi}{2} - \log(2) \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \frac{x}{1 + \cos x} \, dx = \frac{\pi}{2} - \log(2) \]