The value of `int_(0)^(pi//2) (x+sin x)/(1+cos x)dx`, is

Options

A. `pi`
B. `2pi`
C. `pi//2`
D. `3pi//2`

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{2}} \frac{x + \sin x}{1 + \cos x} \, dx \), we will follow a systematic approach.

### Step 1: Rewrite the Integral
We can express the integral as:
\[ I = \int_0^{\frac{\pi}{2}} \frac{x + \sin x}{1 + \cos x} \, dx \]
We can rewrite \( \sin x \) and \( 1 + \cos x \) in terms of half-angle identities:
\[ \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}, \quad 1 + \cos x = 2 \cos^2 \frac{x}{2} \]
Thus, we can rewrite our integral:
\[ I = \int_0^{\frac{\pi}{2}} \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \, dx \]
This simplifies to:
\[ I = \frac{1}{2} \int_0^{\frac{\pi}{2}} \left( \frac{x}{\cos^2 \frac{x}{2}} + 2 \tan \frac{x}{2} \right) dx \]

### Step 2: Split the Integral
Now, we can split the integral into two parts:
\[ I = \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{x}{\cos^2 \frac{x}{2}} \, dx + \int_0^{\frac{\pi}{2}} \tan \frac{x}{2} \, dx \]

### Step 3: Integrate the First Part using Integration by Parts
Let:
\[ u = x, \quad dv = \sec^2 \frac{x}{2} \, dx \]
Then:
\[ du = dx, \quad v = 2 \tan \frac{x}{2} \]
Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ \int_0^{\frac{\pi}{2}} x \sec^2 \frac{x}{2} \, dx = \left[ 2x \tan \frac{x}{2} \right]_0^{\frac{\pi}{2}} - 2 \int_0^{\frac{\pi}{2}} \tan \frac{x}{2} \, dx \]

### Step 4: Evaluate the Limits
Evaluating \( \left[ 2x \tan \frac{x}{2} \right]_0^{\frac{\pi}{2}} \):
- At \( x = \frac{\pi}{2} \):
\[ 2 \cdot \frac{\pi}{2} \cdot \tan \frac{\pi}{4} = \pi \cdot 1 = \pi \]
- At \( x = 0 \):
\[ 2 \cdot 0 \cdot \tan 0 = 0 \]
Thus, the evaluation gives:
\[ \pi - 2 \int_0^{\frac{\pi}{2}} \tan \frac{x}{2} \, dx \]

### Step 5: Combine the Results
Now we have:
\[ I = \frac{1}{2} \left( \pi - 2 \int_0^{\frac{\pi}{2}} \tan \frac{x}{2} \, dx \right) + \int_0^{\frac{\pi}{2}} \tan \frac{x}{2} \, dx \]
This simplifies to:
\[ I = \frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \tan \frac{x}{2} \, dx + \int_0^{\frac{\pi}{2}} \tan \frac{x}{2} \, dx = \frac{\pi}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\frac{\pi}{2}} \]