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

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 will follow these steps:

### Step 1: Rewrite the Integral
We can rewrite the integral as:

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

### Step 2: Use a Trigonometric Identity
Recall the identity for cosine:

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

Substituting this into the integral gives:

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

### Step 3: Change of Variables
Let \( t = \tan\left(\frac{x}{2}\right) \). Then, we have:

\[ dx = \frac{2}{1 + t^2} dt \]

Also, when \( x = 0 \), \( t = 0 \) and when \( x = \frac{\pi}{2} \), \( t = 1 \).

Substituting these into the integral gives:

\[ I = \int_{0}^{1} \frac{\frac{2}{1+t^2} dt}{1 + 2\left(\frac{1-t^2}{1+t^2}\right)} = \int_{0}^{1} \frac{2 dt}{(1+t^2)(1 + 2\frac{1-t^2}{1+t^2})} \]

### Step 4: Simplify the Denominator
Simplifying the denominator:

\[ 1 + 2\frac{1-t^2}{1+t^2} = \frac{(1+t^2) + 2(1-t^2)}{1+t^2} = \frac{3 - t^2}{1+t^2} \]

Thus, the integral becomes:

\[ I = \int_{0}^{1} \frac{2(1+t^2) dt}{(1+t^2)(3-t^2)} = \int_{0}^{1} \frac{2 dt}{3 - t^2} \]

### Step 5: Evaluate the Integral
Now, we can evaluate the integral:

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

Using the formula for the integral of \( \frac{1}{a^2 - x^2} \):

\[ \int \frac{dx}{a^2 - x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{\sqrt{a^2 - x^2}}\right) + C \]

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:

\[ = \frac{2}{\sqrt{3}} \left( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) - \tan^{-1}(0) \right) = \frac{2}{\sqrt{3}} \cdot \frac{\pi}{6} \]

### Step 6: Final Result
Thus, we have:

\[ I = \frac{\pi}{3\sqrt{3}} \]