Find the integral of the function $ \dfrac{1-\cos x}{1+\cos x} $

Hint: Remember the Trigonometric Identities ! There’s always a hint in the question if you observe it carefully.
Trigonometric identities that might be helpful are $ \cos 2\theta =2{{\cos }^{2}}\theta -1 $ , $ \cos 2\theta =1-2{{\sin }^{2}}\theta $ etc.

Complete step-by-step answer:
Trigonometric identities used are:
 $ \cos 2\theta =2{{\cos }^{2}}\theta -1 $ ,
 $ \cos 2\theta =1-2{{\sin }^{2}}\theta $ ,
 $ {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1 $ .
We need to find $ \int{\dfrac{1-\cos x}{1+\cos x}} $ .
We know that $ \cos 2\theta =2{{\cos }^{2}}\theta -1 $ (trigonometric identity)
 $ \cos 2\theta +1=2{{\cos }^{2}}\theta $
 $ {{\cos }^{2}}\theta =\dfrac{\cos 2\theta +1}{2} $
Replacing $ \theta $ with $ \dfrac{x}{2} $ , we get
  $ {{\cos }^{2}}\dfrac{x}{2}=\dfrac{\cos 2\dfrac{x}{2}+1}{2} $
  $ {{\cos }^{2}}\dfrac{x}{2}=\dfrac{\cos x+1}{2} $
  $ 2{{\cos }^{2}}\dfrac{x}{2}=\cos x+1 $ --equation 1
Similarly, we know that $ \cos 2\theta =1-2{{\sin }^{2}}\theta $ (trigonometric identity)
$ 2{{\sin }^{2}}\theta =1-\cos 2\theta $
Replacing $ \theta $ with $ \dfrac{x}{2} $ , we get
 $ 2{{\sin }^{2}}\dfrac{x}{2}=1-\cos x $ --equation 2
Substituting equation 1 and equation 2 , we get
 $\Rightarrow \int{\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}}} $
 $\Rightarrow \int{{{\tan }^{2}}\dfrac{x}{2}} $
We know that $ {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1 $ (trigonometric identity)
Replacing $ \theta $ with $ \dfrac{x}{2} $ , we get
 $ \Rightarrow {{\tan }^{2}}\dfrac{x}{2}={{\sec }^{2}}\dfrac{x}{2}-1 $
Therefore,
 $\Rightarrow \int{\left( {{\sec }^{2}}\dfrac{x}{2}-1 \right)}dx $
 $\Rightarrow \int{{{\sec }^{2}}\dfrac{x}{2}}dx $ $ -\int{dx} $ --equation 3
Let’s find $ \int{{{\sec }^{2}}\dfrac{x}{2}}dx $ by substitution method:
Let $ \dfrac{x}{2} $ be u, then on differentiating both sides we get, $ \dfrac{1}{2}dx=du $
 $ dx=2du $
 $\Rightarrow \int{{{\sec }^{2}}\dfrac{x}{2}}dx=2\int{{{\sec }^{2}}u}du $
We know that $ \int{{{\sec }^{2}}xdx=\tan x} $ +C (trigonometric identity)
Where C is the integral constant
Therefore, $ 2\int{{{\sec }^{2}}u}du $ = $ 2\left( \tan u+C \right) $
Replacing u with $ \dfrac{x}{2} $ , we get
$\Rightarrow 2\tan \dfrac{x}{2}+C $ --equation 4
(2C is also a constant represented by C)
We know that, $ \int{dx} $ = x+C --equation 5
(C being the integral constant)
Substituting equation 4 and equation 5 in equation 3, we get
 $\Rightarrow 2\tan \dfrac{x}{2}-x+C $
Hence, $ \int{\dfrac{1-\cos x}{1+\cos x}} $ = $ 2\tan \dfrac{x}{2}-x+C $
So, the correct answer is “ $ \int{\dfrac{1-\cos x}{1+\cos x}} $ = $ 2\tan \dfrac{x}{2}-x+C $ ”.

Note: Keeping in mind some standard integral rules can save time and solve the question much faster. For example, in this question one need not show integration of $ {{\sec }^{2}}x $ and can directly solve the question. This will take less time to solve the question. Similarly one can remember other standard integral rules, as many as possible!