Question

Integrate \[\dfrac{1}{1+\cot x}\].

Answer 1

Hint: In the type of question that has been stated above first we will be changing \[\cot x\] to \[\sin x\text{ and }\cos x\] and then make the whole equation in terms of sin and cos then we will integrate the new equation that is formed.

Complete step by step answer:
In the above stated question we are going to change \[\cot x\] to \[\sin x\text{ and }\cos x\] as we know that \[\cot x=\dfrac{\cos x}{\sin x}\] by using this relation we can get and equation with respect to sin and cos, now we will substitute this relation of \[\cot x\] into the main equation provided and we get it as:
\[\begin{align}
  & =\dfrac{1}{1+\dfrac{\cos x}{\sin x}} \\
 & =\dfrac{1}{\dfrac{\cos x+\sin x}{\sin x}} \\
 & =\dfrac{\sin x}{\cos x+\sin x} \\
\end{align}\]
Now we will multiply and divide by 2 and then we are going to get
\[=\dfrac{2\sin x}{2\left( \cos x+\sin x \right)}\]
Now in the above statement we are going to add and subtract \[\cos x\] in the numerator with which we get it as:
\[\begin{align}
  & =\dfrac{\sin x+\sin x+\cos x-\cos x}{2\left( \cos x+\sin x \right)} \\
 & =\dfrac{\left( \sin x+\cos x \right)+\left( \sin x-\cos x \right)}{2\left( \cos x+\sin x \right)} \\
 & =\dfrac{1}{2}+\dfrac{\sin x-\cos x}{\sin x+\cos x} \\
\end{align}\]
Now we will integrate the final function that we achieved just now, in this equation we will assume the denominator as t which will when differentiate will be equal to the numerator then the whole 2nd function will change into t and we will get:
\[\begin{align}
  & =\int{\left( \dfrac{1}{2}+\dfrac{\sin x-\cos x}{\sin x+\cos x} \right)}dx \\
 & =\int{\dfrac{1}{2}}dx+\int{\left( \dfrac{\sin x-\cos x}{\sin x+\cos x} \right)dx} \\
\end{align}\]
Let,
 \[\begin{align}
  & t=\sin x+\cos x \\
 & dt=\left( \sin x-\cos x \right)dx \\
\end{align}\]
\[\begin{align}
 & =\dfrac{x}{2}+\int{\dfrac{1}{t}}dt+C \\
 & =\dfrac{x}{2}+\log \left| t \right|+D \\
 & =\dfrac{x}{2}+\log \left| \sin x+\cos x \right|+D \\
\end{align}\]
We substitute the denominator by t and dt by \[\left( \sin x-\cos x \right)dx\] which will give us final equation as \[\dfrac{1}{t}\] and we already know that the integral value of \[\dfrac{1}{t}\] is \[{{\log }_{e}}(t)\]which resulted us with \[\log \left| \sin x+\cos x \right|\], C is the constant produced with the first part of the integral that is the constant i.e. \[\dfrac{1}{2}\]and there will also be another constant produced while integrating another part of the interval i.e. \[\dfrac{1}{t}\] and by adding both the interval we will get a new interval D.

So we get the integral value of the equation mentioned in the question i.e. \[\dfrac{1}{1+\cot x}\] is \[\dfrac{x}{2}+\log \left| \sin x+\cos x \right|+D\].

Note: In the above mentioned question it becomes very easy if we can change the whole equation in sin(x) and cos(x) as it will be very easy to substitute and the derivative of the assumed function is also easily found which will help us to find the final integral value of the desired function.