The integral $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{dx}}{{1 + \cos x}}} $is equal to:
1. $2$
2. $4$
3. $ - 1$
4. $ - 2$

Hint: Here, in this question we are given an integral $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{dx}}{{1 + \cos x}}} $ and we have to find its value. First step is to use trigonometric formula $1 + \cos 2A = 2{\cos ^2}A$ in the denominator and you will get the direct function to integrate i.e., $\int {{{\sec }^2}xdx = } \tan x$. Now to find the values use $\tan \theta = \sqrt {\dfrac{{1 - \cos 2\theta }}{{1 + \cos 2\theta }}} $ and solve further.

Formula Used:
Trigonometric formulas –
$\tan \theta = \sqrt {\dfrac{{1 - \cos 2\theta }}{{1 + \cos 2\theta }}} $
$1 + \cos 2A = 2{\cos ^2}A$
Integration formula –
$\int {{{\sec }^2}xdx = } \tan x$

Complete step by step Solution:
Given that,
$\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{dx}}{{1 + \cos x}}} $
Assume the given integral be $I$,
$I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{dx}}{{1 + \cos x}}} $
$ = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{dx}}{{2{{\cos }^2}\dfrac{x}{2}}}} $
$ = \dfrac{1}{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {{{\sec }^2}\dfrac{x}{2}} dx$
$ = \dfrac{1}{2}\left[ {\dfrac{{\tan \dfrac{x}{2}}}{{\dfrac{1}{2}}}} \right]_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}}$
$ = \tan \dfrac{{3\pi }}{8} - \tan \dfrac{\pi }{8} - - - - - (1)$
Using trigonometric formula, $\tan \theta = \sqrt {\dfrac{{1 - \cos 2\theta }}{{1 + \cos 2\theta }}} $to find the value of $\tan \dfrac{{3\pi }}{8}$and $\tan \dfrac{\pi }{8}$
\[\tan \dfrac{\pi }{8} = \sqrt {\dfrac{{1 - \cos \dfrac{\pi }{4}}}{{1 + \cos \dfrac{\pi }{4}}}} \]
\[ = \sqrt {\dfrac{{1 - \dfrac{1}{{\sqrt 2 }}}}{{1 + \dfrac{1}{{\sqrt 2 }}}}} \]
\[ = \sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \]
\[ = \sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \times \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 - 1}}} \]
Multiply denominator and numerator by \[\sqrt 2 - 1\],
\[ = \dfrac{{\sqrt 2 - 1}}{1}\]
Similarly, \[\tan \dfrac{{3\pi }}{8} = \sqrt 2 + 1\]
Putting the required values in equation (1),
\[ = \left( {\sqrt 2 + 1} \right) - \left( {\sqrt 2 - 1} \right)\]
\[ = 2\]

Hence, the correct option is 1.

Note: To solve such problems one should have a good knowledge of trigonometric and integration formulas. Also, when there’s a function inside the function always apply the chain rule like we do in differentiation after doing the differentiation of the whole function we multiply the differentiation of the inner function. But in this, we integrate the function and then divide the required term by the differentiation of the inner part. It can be written as $\int {f\left( {g\left( x \right)} \right)dx = \dfrac{{\int {f\left( {g\left( x \right)} \right)dx} }}{{\dfrac{d}{{dx}}g\left( x \right)}}} $. Now, to solve the limits in integration. First, integrate the whole function then subtract the integration using the lower limit from the integration using the upper limit. For example, the required integration is $2x$and we have the limit $2$ to $4$ then the answer will be $2\left( 4 \right) - 2\left( 2 \right) = 4$.