`int_(0)^(pi//2)(dx)/(2+cosx)=`

Options

A. `1/(sqrt(3))tan^(-1)(1/(sqrt(3)))`
B. `sqrt(3)tan^(-1)(sqrt(3))`
C. `2/(sqrt(3))tan^(-1)(1/(sqrt(3)))`
D. `2sqrt(3)tan^(-1)(sqrt(3))`

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{2}} \frac{dx}{2 + \cos x} \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_0^{\frac{\pi}{2}} \frac{dx}{2 + \cos x} \]

### Step 2: Use the Weierstrass Substitution
We can use the Weierstrass substitution, which is:
\[ t = \tan\left(\frac{x}{2}\right) \]
This gives us:
\[ \cos x = \frac{1 - t^2}{1 + t^2} \quad \text{and} \quad dx = \frac{2}{1 + t^2} dt \]
The limits change as follows:
- When \( x = 0 \), \( t = \tan(0) = 0 \)
- When \( x = \frac{\pi}{2} \), \( t = \tan\left(\frac{\pi}{4}\right) = 1 \)

### Step 3: Substitute into the Integral
Substituting these into the integral, we have:
\[ I = \int_0^1 \frac{\frac{2}{1 + t^2} dt}{2 + \frac{1 - t^2}{1 + t^2}} \]
Simplifying the denominator:
\[ 2 + \frac{1 - t^2}{1 + t^2} = \frac{2(1 + t^2) + 1 - t^2}{1 + t^2} = \frac{2 + t^2 + 1}{1 + t^2} = \frac{3 + t^2}{1 + t^2} \]
Thus, the integral becomes:
\[ I = \int_0^1 \frac{2}{1 + t^2} \cdot \frac{1 + t^2}{3 + t^2} dt = \int_0^1 \frac{2}{3 + t^2} dt \]

### Step 4: Solve the Integral
Now we need to evaluate:
\[ I = 2 \int_0^1 \frac{1}{3 + t^2} dt \]
Using the formula for the integral of \( \frac{1}{a^2 + x^2} \):
\[ \int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \]
In our case, \( a^2 = 3 \) so \( a = \sqrt{3} \):
\[ I = 2 \cdot \frac{1}{\sqrt{3}} \left[ \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) \right]_0^1 \]
Calculating the limits:
\[ = 2 \cdot \frac{1}{\sqrt{3}} \left( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) - \tan^{-1}(0) \right) \]
Since \( \tan^{-1}(0) = 0 \) and \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \):
\[ I = 2 \cdot \frac{1}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{3\sqrt{3}} \]

### Final Result
Thus, the value of the integral is:
\[ \boxed{\frac{\pi}{3\sqrt{3}}} \]