Question

Integrate : $\int{x{{\cot }^{-1}}(x)dx}$

Answer 1

Hint: Apply Integration By Parts i.e. $\int{f(x)g(x)dx=f(x)\int{g(x)dx}-\int{\{\dfrac{d(f(x))}{dx}*[\int{g(x)dx]\}dx}}}$ where $f(x)$ and $g(x)$ are functions of $x$

Complete step by step answer:
Geometrical meaning of integration is area under the curve when it is drawn in $X-Y$ plane i.e.
$\int\limits_{a}^{b}{Q(x)}dx$ is equal to area under the curve which is enclosed in $x=a$ and $x=b$
Likewise there is another term i.e. Differentiation which is opposite of integration.
Geometrical meaning of Differentiation is that it provides the slope drawn on $y=f(x)$ at point $x=a$ i.e.
$\dfrac{d\{f(x)\}}{dx}$ value at$x=a$ is a slope of tangent drawn at point$\{a,f(a)\}$.
When we have to find the integration of multiplication of two functions we integration by parts i.e. $\int{f(x)g(x)dx=f(x)\int{g(x)dx}-\int{\{\dfrac{d(f(x))}{dx}*[\int{g(x)dx]\}dx}}}$
Firstly before applying integration by parts we first define ${{1}^{st}}$ function and ${{2}^{nd}}$ function according to ILATE rule the function comes first in the word ILATE will be treated as ${{1}^{st}}$function and remaining function as ${{2}^{nd}}$
ILATE stands for I = Inverse Trigonometric function
L = Logarithm function
A = Algebraic function
T= Trigonometric function
E= Exponential function
Let us assume ${{1}^{st}}$ function be $f(x)$ and ${{2}^{nd}}$ function be $g(x)$ then,
\[\int{f(x)g(x)dx=\int{{{1}^{st}}*{{2}^{nd}}dx={{1}^{st}}\int{{{2}^{nd}}dx-\int{\{\dfrac{d({{1}^{st}})}{dx}*\int{({{2}^{nd}})dx}}}}}\}dx\]
We have $\int{x{{\cot }^{-1}}(x)dx}$ ,So first we ${{1}^{st}}$ and ${{2}^{nd}}$ function according to ILATE rule ${{\cot }^{-1}}(x)$ be the ${{1}^{st}}$ function and ${{2}^{nd}}$ function.
So by applying integration by parts we have,
$\int{x*{{\cot }^{-1}}(x)dx={{\cot }^{-1}}(x)\int{xdx-\int{\{\dfrac{d({{\cot }^{-1}}(x)}{dx}*\int{xdx\}dx}}}}..................eq(1)$
$\int{xdx=\dfrac{{{x}^{2}}}{2}+c}......................eq(2)$
$\dfrac{d({{\cot }^{-1}}x)}{dx}=\dfrac{-1}{1+{{x}^{2}}}..............eq(3)$
Put eq(2) and eq(3) in eq(1) then we get,
\[\begin{align}
  & \int{x*{{\cot }^{-1}}(x)dx={{\cot }^{-1}}(x)}*\dfrac{{{x}^{2}}}{2}-\int{\{(}\dfrac{-1}{1+{{x}^{2}}})*\dfrac{{{x}^{2}}}{2}\}dx \\
 & \int{x*{{\cot }^{-1}}(x)dx={{\cot }^{-1}}(x)}*\dfrac{{{x}^{2}}}{2}+\dfrac{1}{2}\int{(\dfrac{{{x}^{2}}}{1+{{x}^{2}}})dx}..............eq(4) \\
\end{align}\]
For $\int{(\dfrac{{{x}^{2}}}{1+{{x}^{2}}})dx}$ we have to add $\pm 1$ to the numerator, so that we can form partial fractions . Therefore we get,
\[\begin{align}
  & \int{(\dfrac{{{x}^{2}}}{1+{{x}^{2}}})dx=}\int{\dfrac{{{x}^{2}}+1-1}{1+{{x}^{2}}}}dx \\
 & \int{(\dfrac{{{x}^{2}}}{1+{{x}^{2}}})dx=}\int{(\dfrac{{{x}^{2}}+1}{{{x}^{2}}+1}-\dfrac{1}{1+{{x}^{2}}})}dx \\
 & \int{(\dfrac{{{x}^{2}}}{1+{{x}^{2}}})dx=}\int{(1-\dfrac{1}{1+{{x}^{2}}})dx} \\
 & \int{(\dfrac{{{x}^{2}}}{1+{{x}^{2}}})dx=}\int{1dx-\int{\dfrac{1}{1+{{x}^{2}}}dx}} \\
\end{align}\]
As we know \[\int{\dfrac{1}{1+{{x}^{2}}}dx={{\tan }^{-1}}(x)+c}\]
So we get \[\int{(\dfrac{{{x}^{2}}}{1+{{x}^{2}}})dx=}\dfrac{{{x}}}{1}-{{\tan }^{-1}}(x)+c'\] put this result in eq(4)
Therefore ,final result will be
\[\begin{align}
  & \int{x*{{\cot }^{-1}}(x)dx={{\cot }^{-1}}(x)}*\dfrac{{{x}^{2}}}{2}+\dfrac{1}{2}\int{(\dfrac{{{x}^{2}}}{1+{{x}^{2}}})dx}+c \\
 & \int{x*{{\cot }^{-1}}(x)dx={{\cot }^{-1}}(x)}*\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}}}{2}-\dfrac{{{\tan }^{-1}}(x)}{2}+c'+c \\
\end{align}\]

Note: Integration to be learn by heart i.e. integration of $\sec x$ and $\cos ecx$
\[\begin{align}
  & \int{\sec (x)dx=\ln (\sec x+\tan x)+c} \\
 & \int{\cos ec(x)dx=\ln (\cos ecx-\cot x})+c \\
\end{align}\]