Calculate the value of the integral \( \int\limits_{\pi /4}^{3\pi /4} {\frac{{dx}}{{1 + \cos x}}} \)

Finding the value of the integral involves definite integration of a trigonometric integral. We will solve this step by step.

Indefinite Trigonometric Integral Calculation

First, we calculate the indefinite integral \( \int {\frac{{dx}}{{1 + \cos x}}} \).

We use the trigonometric identity \( 1 + \cos x = 2 \cos^2(x/2) \).

 \( \int {\frac{{dx}}{{1 + \cos x}}} = \int {\frac{{dx}}{{2 \cos^2(x/2)}}} \) \( = \frac{1}{2}\int {\sec^2(x/2) dx} \) 

Using the standard integral formula \( \int \sec^2(ax) dx = \frac{1}{a} \tan(ax) + C \) with \( a = 1/2 \), the indefinite integral is:

 \( \frac{1}{2}\int {\sec^2(x/2) dx} = \frac{1}{2} \left( {\frac{1}{{1/2}}\tan(x/2)} \right) + C \) \( = \frac{1}{2} (2\tan(x/2)) + C \) \( = \tan(x/2) + C \) 

So, the indefinite integral of \( \frac{dx}{1 + \cos x} \) is \( \tan(x/2) \).

Performing the Definite Integration

Now we apply the limits of integration from \( \pi/4 \) to \( 3\pi/4 \) to find the value of the integral using the result from the indefinite integration:

 \( \int\limits_{\pi /4}^{3\pi /4} {\frac{{dx}}{{1 + \cos x}}} = \left[ {\tan(x/2)} \right]_{\pi /4}^{3\pi /4} \) \( = \tan\left(\frac{3\pi /4}{2}\right) - \tan\left(\frac{\pi /4}{2}\right) \) \( = \tan\left(\frac{3\pi}{8}\right) - \tan\left(\frac{\pi}{8}\right) \) 

Detailed Steps for Integral Evaluation

To complete the integral evaluation, we need the exact values of \( \tan(\pi/8) \) and \( \tan(3\pi/8) \).

Calculating \( \tan(\pi/8) \):

We use the identity \( \tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta} \). Let \( \theta = \pi/8 \). Then \( 2\theta = \pi/4 \).

 \( \tan(\pi/4) = \frac{2\tan(\pi/8)}{1-\tan^2(\pi/8)} \) 

Since \( \tan(\pi/4) = 1 \), let \( t = \tan(\pi/8) \). The equation becomes:

 \( 1 = \frac{2t}{1-t^2} \) \( 1 - t^2 = 2t \) \( t^2 + 2t - 1 = 0 \) 

This is a quadratic equation for \( t \). Using the quadratic formula \( t = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \):

 \( t = \frac{{ - 2 \pm \sqrt {{2^2} - 4(1)(-1)} }}{{2(1)}} \) \( t = \frac{{ - 2 \pm \sqrt {4 + 4} }}{2} \) \( t = \frac{{ - 2 \pm \sqrt 8 }}{2} \) \( t = \frac{{ - 2 \pm 2\sqrt 2 }}{2} \) \( t = - 1 \pm \sqrt 2 \) 

Since \( \pi/8 \) is in the first quadrant (\( 0 < \pi/8 < \pi/2 \)), \( \tan(\pi/8) \) must be positive. Therefore, we take the positive root:

 \( \tan(\pi/8) = \sqrt 2 - 1 \) 

Calculating \( \tan(3\pi/8) \):

We can use the complementary angle property \( \tan(\pi/2 - \theta) = \cot(\theta) \). Let \( \theta = \pi/8 \). Then \( \pi/2 - \pi/8 = 4\pi/8 - \pi/8 = 3\pi/8 \).

 \( \tan(3\pi/8) = \tan(\pi/2 - \pi/8) \) \( = \cot(\pi/8) \) \( = \frac{1}{{\tan(\pi/8)}} \) 

Substitute the value of \( \tan(\pi/8) \):

 \( \tan(3\pi/8) = \frac{1}{{\sqrt 2 - 1}} \) 

Rationalize the denominator:

 \( \tan(3\pi/8) = \frac{1}{{\sqrt 2 - 1}} \times \frac{{\sqrt 2 + 1}}{{\sqrt 2 + 1}} \) \( = \frac{{\sqrt 2 + 1}}{{{{(\sqrt 2 )}^2} - {1^2}}} \) \( = \frac{{\sqrt 2 + 1}}{{2 - 1}} \) \( = \sqrt 2 + 1 \) 

Final Calculation of the value of the integral

Now, substitute the values of \( \tan(3\pi/8) \) and \( \tan(\pi/8) \) back into the definite integral expression:

 \( \left[ {\tan(x/2)} \right]_{\pi /4}^{3\pi /4} = \tan\left(\frac{3\pi}{8}\right) - \tan\left(\frac{\pi}{8}\right) \) \( = (\sqrt 2 + 1) - (\sqrt 2 - 1) \) \( = \sqrt 2 + 1 - \sqrt 2 + 1 \) \( = 2 \) 

The final value of the integral is 2.