Question

How do you find the antiderivative of $ \dfrac{{\cos (x)}}{{1 - \cos (x)}} $ ?

Answer 1

Hint: As we know that an antiderivative of a function $ f(x) $ is a function whose derivative is equal to $ f(x) $ i.e., if $ F'(x) = f(x) $ then $ F'(x) $ is an antiderivative of $ f(x) $ . To find the antiderivative we often reverse the process of differentiation. The general antiderivative of $ f(x) $ is $ F(x) + c $ , where $ F $ is a differentiable function, which means that to find an antiderivative we have to reverse the process of finding a derivative.

Complete step by step solution:
Anti-derivative or differentiation is a reverse operation of differentiation. Otherwise, an anti-derivative is basically an integral, which is the second main concept of calculus.
A function \[F\] is called an antiderivative of \[f\] on an interval \[I\] if \[F'(x) = f(x)\] for all \[x\] in \[I\] .
Consider the given function
   $ \Rightarrow \,\,\,\dfrac{{\cos (x)}}{{1 - \cos (x)}} $
It can rewrite the integral in the simpler form which gives:
 which means $ I = \int {\dfrac{{\cos (x)}}{{1 - \cos (x)}}} \,dx $ ----------(1)
add and subtract 1 in numerator, then
\[ \Rightarrow I = \int {\dfrac{{\cos (x) - 1 + 1}}{{1 - \cos (x)}}} \,dx\]
\[ \Rightarrow I = \int {\dfrac{{\left\{ {\cos (x) - 1} \right\} + 1}}{{1 - \cos (x)}}} \,dx\]
Or separate the integral as and taking the negative sign out, then
\[ \Rightarrow \int {\dfrac{{ - \{ 1 - \cos (x)\} }}{{1 - \cos (x)}}} \,dx + \int {\dfrac{1}{{1 - \cos (x)}}dx} \]
On cancelling the like terms in both numerator and denominator in first integral, we have
 $ \Rightarrow \,I = - \int {dx + \int {\dfrac{{dx}}{{1 - \cos (x)}}} } $
On integrating the first term, we get
 $ I = - x + \int {\dfrac{{dx}}{{1 - \cos (x)}}} $ .---------(2)
For the rest of the remaining integral we will use the tangent half angle substitution where we use $ t = \tan \left( {\dfrac{x}{2}} \right) $ , So the function $ \cos (x) $ can also be expressed in terms of $ \tan \left( {\dfrac{x}{2}} \right) $ in the following ways: $ \cos (x) = \dfrac{{1 - {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}{{1 + {{\tan }^2}\left( {\dfrac{x}{2}} \right)}} $
As we know the trigonometric identity $ 1 + {\tan ^2}x = {\sec ^2}x $ , then
 $ \Rightarrow \cos (x) = \dfrac{{1 - {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}{{{{\sec }^2}\left( {\dfrac{x}{2}} \right)}} $ ---------(3)
Now substituting equation (3) in (2)
\[ \Rightarrow \,\,I = - x + \smallint \dfrac{{dx}}{{\left( {1 - \dfrac{{1 - {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}{{{{\sec }^2}\left( {\dfrac{x}{2}} \right)}}} \right)}}\]
or
\[ \Rightarrow \,I = - x + \int {\dfrac{{{{\sec }^2}\left( {\dfrac{x}{2}} \right)dx}}{{{{\sec }^2}\left( {\dfrac{x}{2}} \right) - \left( {1 - {{\tan }^2}\left( {\dfrac{x}{2}} \right)} \right)}}} \]
Again, by the identity: $ 1 + {\tan ^2}x = {\sec ^2}x $
\[ \Rightarrow \,I = - x + \int {\dfrac{{{{\sec }^2}\left( {\dfrac{x}{2}} \right)dx}}{{1 + {{\tan }^2}\left( {\dfrac{x}{2}} \right) - 1 + {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}} \]
 $ \Rightarrow \,\,I = - x + \int {\dfrac{{{{\sec }^2}\left( {\dfrac{x}{2}} \right)dx}}{{2{{\tan }^2}\left( {\dfrac{x}{2}} \right)}}} $
Multiply and divide 2 in numerator, then
 $ \Rightarrow \,I = - x + \int {\dfrac{{\dfrac{2}{2}{{\sec }^2}\left( {\dfrac{x}{2}} \right)dx}}{{2{{\tan }^2}\left( {\dfrac{x}{2}} \right)}}} $ -------(4)
Now, we should keep that $ t = \tan \dfrac{x}{2} $ , then
 $ \Rightarrow \,\,dt = \dfrac{1}{2}{\sec ^2}\left( {\dfrac{x}{2}} \right)dx $ -------(5)
by substituting the equation (5) in (4), then
  $ \Rightarrow \,\,\,I = - x + \int {\dfrac{{dt}}{{{t^2}}}} $
\[ \Rightarrow \,\,I = - x - \dfrac{1}{t} + C\]
Substitute the t value
 $ \Rightarrow \,\,I = - x - \dfrac{1}{{\tan \left( {\dfrac{x}{2}} \right)}} + C $
As we know the reciprocal of tan is cot, then
  $ I = - x - \cot \left( {\dfrac{x}{2}} \right) + C $ .
Hence the required answer is $ - x - \cot \left( {\dfrac{x}{2}} \right) + C $ .
So, the correct answer is “ $ - x - \cot \left( {\dfrac{x}{2}} \right) + C $ ”.

Note: We should always keep in mind that $ c - c \ne 0 $ , because there are constants and we do not know what another number is there in our antiderivative. Infact $ c - c $ will always be a constant and since $ c $ represents a constant, we can just call it normal $ c $ . While calculating antiderivative we should never forget $ C $ as our final answer always has it.