Question

Integrate $\dfrac{1}{1-\cot x}$ with respect to x.

Answer 1

In this question, we have to find the value of $\int{\dfrac{1}{1-\cot x}dx.}$ For this, we will use the following properties
$\left( i \right)\cot x=\dfrac{\cos x}{\sin x}$
$\left( ii \right)\int{\dfrac{1}{x}dx=\ln \left| x \right|+c}$
$\left( iii \right)\int{1.dx=x}$
We will first simplify our given function and then apply (ii) and (iii) to evaluate our integral.

Complete step-by-step answer:
We are given our integral as $\int{\dfrac{1}{1-\cot x}dx.}$ Let us first simplify the given functions. We know that $\cot x=\dfrac{\cos x}{\sin x},$ so we get
$\int{\dfrac{1}{1-\dfrac{\cos x}{\sin x}}dx}$
Taking LCM, we get,
$\Rightarrow \int{\dfrac{1}{\dfrac{\sin x-\cos x}{\sin x}}dx}$
$\Rightarrow \int{\dfrac{\sin x}{\sin x-\cos x}dx}$
Now, multiplying and dividing by 2, we get,
$\Rightarrow \int{\dfrac{2\sin x}{2\left( \sin x-\cos x \right)}dx}$
$\Rightarrow \dfrac{1}{2}\int{\dfrac{2\sin x}{\left( \sin x-\cos x \right)}dx}$
$\Rightarrow \dfrac{1}{2}\int{\dfrac{\sin x+\sin x}{\sin x-\cos x}dx}$
Adding and subtracting cos x in the numerator, we get,
$\Rightarrow \dfrac{1}{2}\int{\dfrac{\sin x+\sin x+\cos x-\cos x}{\sin x-\cos x}dx}$
Rearranging the terms, we get,
$\Rightarrow \dfrac{1}{2}\int{\left( \dfrac{\sin x-\cos x}{\sin x-\cos x}+\dfrac{\sin x+\cos x}{\sin x-\cos x} \right)dx}$
$\Rightarrow \dfrac{1}{2}\int{\left( 1+\dfrac{\sin x+\cos x}{\sin x-\cos x} \right)dx}$
Separating the integrals, we get,
$\Rightarrow \dfrac{1}{2}\int{1.dx+\dfrac{1}{2}\int{\left( \dfrac{\sin x+\cos x}{\sin x-\cos x} \right)}dx}......\left( i \right)$
Now, let us evaluate the integral separately. Let us assume, ${{I}_{1}}=\int{1.dx}$ and ${{I}_{2}}=\int{\dfrac{\left( \sin x+\cos x \right)}{\left( \sin x-\cos x \right)}dx}.$ Let us evaluate ${{I}_{1}}$ first. So, we have, ${{I}_{1}}=\int{1.dx}.$
We know that, $\int{1.dx}=x+c,$ so we get,
${{I}_{1}}=x+{{c}_{1}}\left[ {{c}_{1}}\text{ is any constant} \right]$
Now, let us evaluate ${{I}_{2}}.$ We have,
${{I}_{2}}=\int{\dfrac{\left( \sin x+\cos x \right)}{\left( \sin x-\cos x \right)}dx}$
Let us put sin x – cos x = t. Taking the derivative with respect to x, we get,
$\Rightarrow \dfrac{d}{dx}\left( \sin x-\cos x \right)=\dfrac{dt}{dx}$
Since, $\dfrac{d}{dx}\sin x=\cos x$ and $\dfrac{d}{dx}\cos x=-\sin x.$ So, we get,
$\Rightarrow \cos x-\left( -\sin x \right)=\dfrac{dt}{dx}$
$\Rightarrow \sin x+\cos x=\dfrac{dt}{dx}$
Cross multiplying we get,
$\Rightarrow dt=\left( \sin x+\cos x \right)dx$
In ${{I}_{2}}$ putting the values, we get,
${{I}_{2}}=\int{\dfrac{dt}{t}}$
$\Rightarrow {{I}_{2}}=\int{\dfrac{1}{t}dt}$
We know that, $\int{\dfrac{1}{x}dx=\ln \left| x \right|+c}.$
Hence, we get,
$\Rightarrow {{I}_{2}}=\ln \left| t \right|+{{c}_{2}}\left[ {{c}_{2}}\text{ is any constant} \right]$
Since t was supposed to be sin x – cos x, so,
$\Rightarrow {{I}_{2}}=\ln \left| \sin x-\cos x \right|+{{c}_{2}}$
Now putting the values of ${{I}_{1}}$ and ${{I}_{2}}$ in (i), we get,
$\Rightarrow \dfrac{1}{2}\left( x+{{c}_{1}} \right)+\dfrac{1}{2}\left( \ln \left| \sin x-\cos x \right|+{{c}_{2}} \right)$
$\Rightarrow \dfrac{x}{2}+\dfrac{{{c}_{1}}}{2}+\dfrac{1}{2}\ln \left| \sin x-\cos x \right|+\dfrac{{{c}_{2}}}{2}$
$\Rightarrow \dfrac{x}{2}+\dfrac{1}{2}\ln \left| \sin x-\cos x \right|+c\left[ c=\dfrac{{{c}_{1}}}{2}+\dfrac{{{c}_{2}}}{2} \right]$
Hence,
$\int{\dfrac{1}{1-\cot x}dx}=\dfrac{x}{2}+\dfrac{1}{2}\ln \left| \sin x-\cos x \right|+c$

Note: The most common mistake that students can make is to forget adding the constant term after evaluating the integral. Take care of the signs while simplifying the fractions. Also, take care of the signs while substituting the values of sin x – cos x. Make sure to take the negative sign with sin x as a derivative of cos x. At the end, we have taken $\dfrac{{{c}_{1}}}{2}+\dfrac{{{c}_{2}}}{2}$ as c because $c,{{c}_{1}},{{c}_{2}}$ are any constants and adding ${{c}_{1}},{{c}_{2}}$ will give any constant which we assumed as c.