Question

How do you find the integral of \[\int{\dfrac{1}{1+\cot x}}\] ?

Answer 1

Hint:In the given question, we have been asked to integrate the given constant. In order to solve the question, we integrate the numerical by using the basic concept of integration. First we need to simplify the given integration. Later we will need to integrate the variable part using a suitable integration formula and we will get our required answer.

Complete step by step answer:
We have given,
\[\Rightarrow \int{\dfrac{1}{1+\cot x}}dx\]
Let I be the integration of the given equation.
Therefore, we can write the integration as,
\[\Rightarrow I=\int{\dfrac{1}{1+\cot x}}dx\]
Simplifying the above expression, we obtained
\[\Rightarrow I=\int{\dfrac{1}{1+\cot x}}dx\\
\Rightarrow I=\int{\dfrac{1}{1+\dfrac{\cos x}{\sin x}}dx=}\int{\dfrac{\sin x}{\sin x+\cos x}}dx\\
\Rightarrow I=\dfrac{1}{2}\int{\dfrac{2\sin x}{\sin x+\cos x}}dx\]
Therefore, we have
\[\Rightarrow I=\dfrac{1}{2}\int{\dfrac{2\sin x}{\sin x+\cos x}}dx\]
Expanding the above expression further, we get
\[\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \sin x+\cos x \right)+\left( \sin x-\cos x \right)}{\left( \sin x+\cos x \right)}}dx\]
\[\Rightarrow I=\dfrac{1}{2}\int{1+\dfrac{\left( \sin x-\cos x \right)}{\left( \sin x+\cos x \right)}}dx\]
Integrating each part individually, we get
\[\Rightarrow I=\dfrac{1}{2}\left( \int{1dx}+\int{\dfrac{\left( \sin x-\cos x \right)}{\left( \sin x+\cos x \right)}dx} \right)\]
Therefore,
Using the integration rules, we have
\[\Rightarrow I=\dfrac{1}{2}x+\dfrac{1}{2}\int{\dfrac{\left( \sin x-\cos x \right)}{\left( \sin x+\cos x \right)}dx}\]
Now,let
\[\Rightarrow t=\left( \sin x+\cos x \right)\\
\Rightarrow dt=\left( \cos x-\sin x \right)dx\]
\[\Rightarrow I=\dfrac{1}{2}x+\dfrac{1}{2}\int{\dfrac{-dt}{t}}\]
Applying the standard formula of integration, we obtained
\[\Rightarrow I=\dfrac{1}{2}x+\dfrac{1}{2}\left( -\ln \left| t \right| \right)+C\]
Therefore,we have
\[\Rightarrow I=\dfrac{x}{2}-\dfrac{\ln \left| t \right|}{2}+C\]
Undo the substitution, we obtained
\[\Rightarrow I=\dfrac{x}{2}-\dfrac{1}{2}\ln \left| \sin x+\cos x \right|+C\]
\[\therefore\int{\dfrac{1}{1+\cot x}}=\dfrac{x}{2}-\dfrac{1}{2}\ln \left| \sin x+\cos x \right|+C\]

Hence,the integral of \[\int{\dfrac{1}{1+\cot x}}\] is $\dfrac{x}{2}-\dfrac{1}{2}\ln \left| \sin x+\cos x \right|+C$.

Note:While solving the questions of integration, students always need to remember that they must not forget to put the constant term C after the integration of an equation. The value of this constant can be anything as it can be any whole number or integer or it can also be zero. The question given above is just a simple integration of a constant for that we should always remember to take out the constant term out of the integration and integrate the remaining integral by using the basic concepts or methods of integration. We should remember the property or the formulas of integration, this would make it easier to solve the question.