Question

Evaluate the following integral $ \int{\dfrac{1}{1-\cos x}}dx $

Answer 1

Hint: For answering this type of questions where the function is in the form of $ \dfrac{1}{a\sin x+b\cos x} $ or $ \dfrac{1}{a\pm b\cos x} $ or $ \dfrac{1}{a\pm b\sin x} $ , we need to convert the terms $ \sin \left( x \right) $ and $ \cos \left( x \right) $ into $ \tan \left( \dfrac{x}{2} \right) $ . Then after doing this conversion we have to do the correct substitutions to get the simplified equation.

Complete step by step answer:
Now considering the given expression in the question we have $ \int{\dfrac{1}{1-\cos x}}dx $ .
The integration of the given expression $ \int{\dfrac{1}{1-\cos x}}dx $ is in the form of $ \dfrac{1}{a\pm b\cos x} $ .
Let us assume $ \tan \dfrac{x}{2}=t $ by differentiating this function we will have $ \dfrac{1}{2}{{\sec }^{2}}\left( \dfrac{x}{2} \right)dx=dt $ using the formulae $ \dfrac{d}{dx}\tan \left( \dfrac{x}{a} \right)=\dfrac{1}{a}{{\sec }^{2}}\left( \dfrac{x}{a} \right) $ .
Let us assume that $ I=\int{\dfrac{1}{1-\cos x}}dx $
Now by substituting the value of $ dx $ here we will have
 $ \begin{align}
  & I=\int{\dfrac{1}{1-\cos x}}\left( \dfrac{dt}{\dfrac{1}{2}{{\sec }^{2}}\left( \dfrac{x}{2} \right)} \right) \\
 & \Rightarrow I=2\int{\dfrac{1}{1-\cos x}\left( {{\cos }^{2}}\left( \dfrac{x}{2} \right) \right)dt} \\
\end{align} $
By further simplifying this we will have $ I=2\int{\dfrac{1}{1-\cos x}\left( {{\cos }^{2}}\left( \dfrac{x}{2} \right) \right)dt} $ using $ \cos x=2{{\cos }^{2}}\left( \dfrac{x}{2} \right)-1 $ we will have
  $ \begin{align}
  & I=2\int{\dfrac{1}{1-\left( 2{{\cos }^{2}}\left( \dfrac{x}{2} \right)-1 \right)}\left( {{\cos }^{2}}\left( \dfrac{x}{2} \right) \right)dt} \\
 & \Rightarrow I=2\int{\dfrac{1}{2-2{{\cos }^{2}}\left( \dfrac{x}{2} \right)}}\left( {{\cos }^{2}}\left( \dfrac{x}{2} \right) \right)dt \\
 & \Rightarrow I=\int{\dfrac{{{\cos }^{2}}\left( \dfrac{x}{2} \right)}{1-{{\cos }^{2}}\left( \dfrac{x}{2} \right)}dt} \\
\end{align} $
Now by dividing both numerator and denominator by $ {{\cos }^{2}}\left( \dfrac{x}{2} \right) $ we will have $ I=\int{\dfrac{1}{{{\sec }^{2}}\left( \dfrac{x}{2} \right)-1}dt} $
Now as we have differentiation in terms of $ t $ then we need to express the function in terms of $ t $ . That is we need to express $ {{\sec }^{2}}\left( \dfrac{x}{2} \right) $ in terms of $ \tan \left( \dfrac{x}{2} \right) $ .
We will use the trigonometric identity $ 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $ . So we can write the function as

 $ I=\int{\dfrac{1}{{{\tan }^{2}}\left( \dfrac{x}{2} \right)}dt} $
 $ \Rightarrow I=\int{\dfrac{1}{{{t}^{2}}}dt} $
By using the formulae $ \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C,\text{when }n\ne 1} $ .
 $ \Rightarrow I=\left( \dfrac{-1}{t} \right)+c $
By substituting the value of $ t=\tan \left( \dfrac{x}{2} \right) $ . We will have
 $ \Rightarrow I=\dfrac{-1}{\tan \left( \dfrac{x}{2} \right)}+c $
Therefore, the simplified form of the given integrated equation is $ I=-\cot \left( \dfrac{x}{2} \right)+c $
Hence, $ \int{\dfrac{1}{1-\cos x}}dx=-\cot \left( \dfrac{x}{2} \right)+c $
Hence the given question is evaluated.

Note:
While answering questions of this type we must be aware of the basic integration formulae which will help us in solving this type of question. We should not neglect the integration constant sometimes it makes a difference in the answer. Most of the students will neglect the integration constant. The calculations must be done very carefully. We should remember the basic formulae $ \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C,\text{when }n\ne -1} $ and $ \int{\dfrac{1}{x}dx=\log x+C} $ .