Question

How do you find \[\int { - \arctan \left( {\cot x} \right)dx?} \]

Answer 1

Hint: In this question we have to find the integral of the tan inverse of cotangent of \[x\] . Firstly, we will simplify the expression using inverse trigonometric identity i.e., \[{\tan ^{ - 1}}\theta + {\cot ^{ - 1}}\theta = \dfrac{\pi }{2}\] . After that we will substitute the value in the given expression and using the formula \[{\cot ^{ - 1}}\left( {\cot \theta } \right) = \theta \] we will further simplify the expression and then solve the integral. Hence, we will get the required result.
Formulas used to solve the integral are: -
(1) \[{\int x ^n}{\text{ }}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\] where \[c\] is the constant of integration
(2) \[\int a {\text{ }}dx = ax + c\] where \[c\] is the constant of integration

Complete step by step answer:
We have to find the integral of the tan inverse of cotangent of \[x\]
i.e., \[\int { - {{\tan }^{ - 1}}\left( {\cot x} \right)dx} {\text{ }} - - - \left( 1 \right)\]
Firstly, we will simplify the expression using inverse trigonometric identity
i.e., \[{\tan ^{ - 1}}\theta + {\cot ^{ - 1}}\theta = \dfrac{\pi }{2}\]
\[ \Rightarrow {\tan ^{ - 1}}\theta = \dfrac{\pi }{2} - {\cot ^{ - 1}}\theta \]
Here, in the question \[\theta = \cot x\]
Therefore, we get
\[{\tan ^{ - 1}}\left( {\cot x} \right) = \dfrac{\pi }{2} - {\cot ^{ - 1}}\left( {\cot x} \right)\]
So, from equation \[\left( 1 \right)\] we have
\[\int { - \left( {\dfrac{\pi }{2} - {{\cot }^{ - 1}}\left( {\cot x} \right)} \right)} {\text{ }}dx{\text{ }} - - - \left( 2 \right)\]
Now, we know that
\[{\cot ^{ - 1}}\left( {\cot \theta } \right) = \theta \]
From equation \[\left( 2 \right)\]
\[\theta = x\]
Therefore, \[{\cot ^{ - 1}}\left( {\cot x} \right) = x - - - \left( 3 \right)\]
Thus, using equation \[\left( 3 \right)\] in equation \[\left( 2 \right)\] we get
\[\int { - \left( {\dfrac{\pi }{2} - x} \right)} {\text{ }}dx\]
On multiplying with the negative sign, we get
\[\int {\left( {x - \dfrac{\pi }{2}} \right)} {\text{ }}dx{\text{ }} - - - \left( 4 \right)\]
Now we know that
\[{\int x ^n}{\text{ }}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\]
And \[\int a {\text{ }}dx = ax + c\]
So, from equation \[\left( 4 \right)\] we get
\[\int {\left( {x - \dfrac{\pi }{2}} \right)} {\text{ }}dx{\text{ = }}\dfrac{{{x^2}}}{{2}} - \dfrac{\pi }{2}x + c\]

Note:
In an indefinite integral, we will put a constant of integration \[c\] not in a definite integral. Alternatively, we can solve this question by another method i.e.,
We have to find the integral of the tan inverse of cotangent of \[x\]
i.e., \[\int { - {{\tan }^{ - 1}}\left( {\cot x} \right)dx} {\text{ }} - - - \left( 1 \right)\]
Firstly, we will simplify the expression using trigonometric identity
i.e., \[\cot x = \tan \left( {\dfrac{\pi }{2} - x} \right)\]
therefore, from the equation \[\left( 1 \right)\] we have
\[\int { - {{\tan }^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{2} - x} \right)} \right)dx} {\text{ }} - - - \left( 2 \right)\]
Now we know that
\[{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta \]
Here, \[\theta = \dfrac{\pi }{2} - x\]
\[ \Rightarrow {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{2} - x} \right)} \right) = \dfrac{\pi }{2} - x\]
Therefore, from the equation \[\left( 2 \right)\] we have
\[\int { - \left( {\dfrac{\pi }{2} - x} \right)dx} {\text{ }} - - - \left( 3 \right)\]
On multiplying with the negative sign, we get
\[\int {\left( {x - \dfrac{\pi }{2}} \right)} {\text{ }}dx{\text{ }} - - - \left( 4 \right)\]
Now we know that
\[{\int x ^n}{\text{ }}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\]
And \[\int a {\text{ }}dx = ax + c\]
So, from equation \[\left( 4 \right)\] we get
\[\int {\left( {x - \dfrac{\pi }{2}} \right)} {\text{ }}dx{\text{ = }}\dfrac{{{x^2}}}{{2}} - \dfrac{\pi }{2}x + c\]