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# Evaluate $\int\limits_0^1 {{{\cot }^{ - 1}}(1 - x + {x^2})dx}$A.$\pi - \dfrac{{\log }}{2}$B. $2\pi - \dfrac{{\log }}{2}$C. $\pi + \dfrac{{\log }}{2}$D. None of these

Last updated date: 02nd Aug 2024
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Hint: We use the property that tangent and cotangent functions are reciprocal of each other. Convert the integral in terms of inverse tangent function and break the numerator in such a way that we can apply the property of addition of two tangent inverse functions. Use integration by parts to find the value of the obtained integral.
* $\because \cot \theta = \dfrac{1}{{\tan \theta }} \Rightarrow {\cot ^{ - 1}}\theta = {\tan ^{ - 1}}\dfrac{1}{\theta }$
* ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)$
*In integration by parts we use the sequence of ILATE (Inverse trigonometric, Logarithms, algebraic, trigonometric, and exponential) for choosing the value of $u,v$ and then substitute in the integration formula of by parts integration.

Complete step by step solution:
We have to evaluate the integral $\int\limits_0^1 {{{\cot }^{ - 1}}(1 - x + {x^2})dx}$
Convert the integral in terms of tangent inverse
$\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}\dfrac{1}{{(1 - x + {x^2})}}} dx$................… (1)
Now we break the numerator in such a way so we can apply property of trigonometry
We can write $1 = x + 1 - x$
Substitute the value of $1 = x + 1 - x$ in numerator of equation (1)
$\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}\dfrac{{x + 1 - x}}{{(1 - x + {x^2})}}} dx$
Write the denominator in easy terms
$\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}\dfrac{{x + 1 - x}}{{1 - x(1 - x)}}} dx$...............… (2)
If we take two angles, $(x)$ and $(1 - x)$ then we can write ${\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(1 - x) = {\tan ^{ - 1}}\left( {\dfrac{{x + (1 - x)}}{{1 - x(1 - x)}}} \right)$
So, RHS of the equation (2) transforms into easier form using the identity ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)$
$\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {\left( {{{\tan }^{ - 1}}(x) + {{\tan }^{ - 1}}(1 - x)} \right)dx}$
Separate the two inverse functions in RHS
$\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}(x)dx} + \int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx}$.................… (3)
Since we know that $\int\limits_0^a {f(x)} dx = \int\limits_0^a {f(a - x)} dx$
So, $\int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} = \int\limits_0^1 {{{\tan }^{ - 1}}1 - (1 - x)dx}$
Which on solving gives $\int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} = \int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - 1 + x} \right)dx}$
i.e. $\int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} = \int\limits_0^1 {{{\tan }^{ - 1}}xdx}$
Substitute the value of $\int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} = \int\limits_0^1 {{{\tan }^{ - 1}}xdx}$ in equation (3)
$\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}(x)dx} + \int\limits_0^1 {{{\tan }^{ - 1}}(x)dx}$
$\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = 2\int\limits_0^1 {{{\tan }^{ - 1}}(x)dx}$
Now we can write RHS as
$\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = 2\int\limits_0^1 {{{\tan }^{ - 1}}x \times 1dx}$
Use by parts to integrate RHS
Looking at the integral $2\int\limits_0^1 {{{\tan }^{ - 1}}x \times 1dx}$
Using the ILATE sequence we can tell that the term comes first in the sequence is $u$ and which comes later is $v$ i.e. in this case we have the terms inverse trigonometric and algebraic , first comes the inverse trigonometric term then comes the algebraic term so $u$ is ${\tan ^{ - 1}}x$ and $v$ is 1
$u = {\tan ^{ - 1}}x,v = 1$
Now we calculate the differentiation of the term $u$
$u' = \dfrac{{du}}{{dx}}$
$\Rightarrow u' = \dfrac{{d({{\tan }^{ - 1}}x)}}{{dx}}$
$\Rightarrow u' = \dfrac{1}{{1 + {x^2}}}$
Now substitute the values in equation $\int {uvdt = u\int {vdt - \int {u'\left( {\int {vdt} } \right)dt} } }$
$\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2{\tan ^{ - 1}}x\int\limits_0^1 {1dx} - 2\int\limits_0^1 {(\dfrac{1}{{1 + {x^2}}})\left( {\int {1dx} } \right)dx}$
Now we know $\int {1dx = x}$
Therefore substitute the value of $\int {1dx = x}$ in RHS
$\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2{\tan ^{ - 1}}x(x) - 2\int\limits_0^1 {(\dfrac{1}{{1 + {x^2}}})(x)dx}$
Now we can solve and bring out constant terms from the integral sign.
$\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2x{\tan ^{ - 1}}x - \int\limits_0^1 {(\dfrac{{2x}}{{1 + {x^2}}})dx}$
Since the numerator of the fraction in RHS is derivative of the denominator and we know $\int {\dfrac{{dt}}{t} = \log t}$
$\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {x{{\tan }^{ - 1}}x} \right]_0^1 - \left[ {\log \left( {1 + {x^2}} \right)} \right]_0^1$
Calculate the values by substituting the upper and lower limits
$\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {1{{\tan }^{ - 1}}1 - 0} \right] - \left[ {\log \left( {1 + 1} \right) - \log \left( {1 + 0} \right)} \right]$
$\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {{{\tan }^{ - 1}}1} \right] - \left[ {\log 2 - \log 1} \right]$
Substitute ${\tan ^{ - 1}}1 = \dfrac{\pi }{4}$ and use the value $\log 1 = 0$
$\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2 \times \dfrac{\pi }{4} - \left[ {\log 2 - 0} \right]$
$\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = \dfrac{\pi }{2} - \log 2$
So value of $2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx}$ is $\dfrac{\pi }{2} - \log 2$.

$\therefore$Option D is correct.

Note: Alternative method:
We can solve the solution after the step $2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {{{\tan }^{ - 1}}1} \right] - \left[ {\log 2 - \log 1} \right]$ in another way
For step in the end we can use rule of log i.e. $\log m - \log n = \log \dfrac{m}{n}$
$\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {{{\tan }^{ - 1}}1} \right] - \left[ {\log \dfrac{2}{1}} \right]$
$\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = \dfrac{\pi }{2} - \log 2$
So value of $2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx}$is$\dfrac{\pi }{2} - \log 2$.
$\therefore$Option D is correct.