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.