Choose the correct Answer of the Following Questions : `int_0^(pi/2) dx /(2+cosx)=`

Options

A. `1/3 tan^(-1)(1/sqrt3)`
B. `2/3 tan^(-1)(1/sqrt3)`
C. 1/2
D. 2

Answer

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

### Step 1: Use the substitution for cosine
We can use the substitution \( \cos x = 1 - 2 \sin^2\left(\frac{x}{2}\right) \) or the half-angle formula. However, a more straightforward approach is to use the substitution \( t = \tan\left(\frac{x}{2}\right) \). This gives us:
\[ \cos x = \frac{1 - t^2}{1 + t^2} \]
and the differential \( dx = \frac{2}{1 + t^2} dt \).

### Step 2: Change the limits of integration
When \( x = 0 \), \( t = \tan(0) = 0 \). When \( x = \frac{\pi}{2} \), \( t = \tan\left(\frac{\pi}{4}\right) = 1 \). Thus, the limits of integration change from \( 0 \) to \( 1 \).

### Step 3: Substitute into the integral
Now substituting into the integral, we have:
\[ \int_0^{\frac{\pi}{2}} \frac{dx}{2 + \cos x} = \int_0^1 \frac{\frac{2}{1 + t^2} dt}{2 + \frac{1 - t^2}{1 + t^2}} \]
This simplifies to:
\[ \int_0^1 \frac{2}{1 + t^2} \cdot \frac{1 + t^2}{2(1 + t^2) + (1 - t^2)} dt = \int_0^1 \frac{2}{3 + t^2} dt \]

### Step 4: Evaluate the integral
Now we need to evaluate:
\[ \int_0^1 \frac{2}{3 + t^2} dt \]
This can be solved using the formula for the integral of the form \( \int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \).

Here, \( a^2 = 3 \) so \( a = \sqrt{3} \):
\[ \int \frac{2}{3 + t^2} dt = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) \]

### Step 5: Apply the limits
Now we apply the limits from \( 0 \) to \( 1 \):
\[ \left[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) \right]_0^1 = \frac{2}{\sqrt{3}} \left( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) - \tan^{-1}(0) \right) \]
Since \( \tan^{-1}(0) = 0 \) and \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \):
\[ = \frac{2}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{3\sqrt{3}} \]

### Final Answer
Thus, the value of the definite integral \( \int_0^{\frac{\pi}{2}} \frac{dx}{2 + \cos x} \) is:
\[ \frac{\pi}{3\sqrt{3}} \]