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

Options

A. `(pi)/(3)`
B. `(pi)/(3sqrt(3))`
C. `(pi)/(sqrt(3))`
D. `(pi)/(2sqrt(3))`

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{2 + \cos x} \, dx, \]

we can use a trigonometric substitution.

### Step 1: Rewrite the cosine function
We can express \(\cos x\) in terms of the tangent function. Using the identity:

\[ \cos x = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}, \]

we substitute this into the integral.

### Step 2: Substitute into the integral
Substituting \(\cos x\) gives us:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{2 + \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}} \, dx. \]

This simplifies to:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2\left(\frac{x}{2}\right)}{2(1 + \tan^2\left(\frac{x}{2}\right)) + 1 - \tan^2\left(\frac{x}{2}\right)} \, dx. \]

### Step 3: Simplify the expression
The denominator simplifies to:

\[ 2(1 + \tan^2\left(\frac{x}{2}\right)) + 1 - \tan^2\left(\frac{x}{2}\right) = 3 + \tan^2\left(\frac{x}{2}\right). \]

Thus, we have:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2\left(\frac{x}{2}\right)}{3 + \tan^2\left(\frac{x}{2}\right)} \, dx. \]

### Step 4: Change of variable
Let \( t = \tan\left(\frac{x}{2}\right) \). Then, \( dx = \frac{2}{1 + t^2} dt \). The limits change from \( x = 0 \) to \( x = \frac{\pi}{2} \) which corresponds to \( t = 0 \) to \( t = 1 \).

### Step 5: Substitute and integrate
Now substituting \( dx \) and changing the limits gives:

\[ I = \int_{0}^{1} \frac{1 + t^2}{3 + t^2} \cdot \frac{2}{1 + t^2} dt = 2 \int_{0}^{1} \frac{1}{3 + t^2} dt. \]

### Step 6: Evaluate the integral
Now we can evaluate:

\[ \int \frac{1}{3 + t^2} dt = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right). \]

Thus,

\[ I = 2 \cdot \left[ \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) \right]_{0}^{1} = 2 \cdot \frac{1}{\sqrt{3}} \left( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) - \tan^{-1}(0) \right). \]

### Step 7: Final evaluation
Since \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\) and \(\tan^{-1}(0) = 0\), we have:

\[ I = 2 \cdot \frac{1}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{3\sqrt{3}}. \]

### Final Answer
Thus, the final result is:

\[ \int_{0}^{\frac{\pi}{2}} \frac{1}{2 + \cos x} \, dx = \frac{\pi}{3\sqrt{3}}. \]