Question

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

Answer 1

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
$\int {\left( {x - \dfrac{\pi }{2}} \right)} {\text{ }}dx{\text{ = }}\dfrac{{{x^2}}}{{2}} - \dfrac{\pi }{2}x + c$