Question

Find the Integration of - $\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 + x + {x^2}} \right)dx} $

Answer 1

Hint: To solve this question first we use ( ${\cot ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{1}{x}$) and then using the formula of inverse trigonometric function (${\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}x - {\tan ^{ - 1}}y$) using this property we can easily solve the question further.

Complete step-by-step answer:
We have given in the question:
$\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 + x + {x^2}} \right)dx} $
Now using ${\cot ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{1}{x}$ this property we can write it as
$\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\dfrac{1}{{1 + x + {x^2}}}} \right)dx} $
We can write it as also
$\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\dfrac{1}{{1 + x\left( {1 + x} \right)}}} \right)dx} $
To apply property we will write it as :
$\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\dfrac{{\left( {1 + x} \right) - x}}{{1 + x\left( {1 + x} \right)}}} \right)dx} $
Now we can apply the property ${\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}x - {\tan ^{ - 1}}y$ using this property we can break it into two parts.
$\int\limits_0^1 {\left[ {\left( {{{\tan }^{ - 1}}\left( {1 + x} \right)} \right) - {{\tan }^{ - 1}}x} \right]dx} $
Now for further integration we will use integration by parts:
We will write it as
$\int\limits_0^1 {1.{{\tan }^{ - 1}}\left( {1 + x} \right)dx} - \int\limits_0^1 {1.{{\tan }^{ - 1}}\left( x \right)dx} $
${\left[ {x{{\tan }^{ - 1}}\left( {x + 1} \right)} \right]^1}_0 - \int\limits_0^1 {\dfrac{x}{{1 + {x^2}}}dx} - {\left[ {x{{\tan }^{ - 1}}x - \int\limits_0^1 {\dfrac{1}{{{x^2}}}.xdx} } \right]^1}_0$
${\left[ {x{{\tan }^{ - 1}}\left( {x + 1} \right)} \right]^1}_0 - \int\limits_0^1 {\dfrac{x}{{1 + {x^2}}}dx} - {\left[ {x{{\tan }^{ - 1}}x - \int\limits_0^1 {\dfrac{{dx}}{x}} } \right]^1}_0$
Now on integrating we get,
${\left[ {x{{\tan }^{ - 1}}\left( {x + 1} \right)} \right]^1}_0 - {\left[ {\dfrac{1}{2}\log \left( {1 + {x^2}} \right)} \right]^1}_0 - {\left[ {x{{\tan }^{ - 1}}x} \right]^1}_0 + {\left[ {\log x} \right]^1}_0$
On putting the limit we get,
${\tan ^{ - 1}}2 - \dfrac{1}{2}\log 2 - \dfrac{\pi }{4}$

Note:- Whenever we get this type of question the key concept of solving is we have to have knowledge of integration and remember formula of integration by parts that is $\int {uvdx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)dx} } } $
Because this is the most important method for integration.