Question

The value of the integral $\int\limits_{0}^{1}{x{{\cot }^{-1}}\left( 1-{{x}^{2}}+{{x}^{4}} \right)dx}$ is
A. $\dfrac{\pi }{4}-\dfrac{1}{2}{{\log }_{e}}2$
B. $\dfrac{\pi }{2}{{\log }_{e}}2$
C. $\dfrac{\pi }{2}-\dfrac{1}{2}{{\log }_{e}}2$
D. $\dfrac{\pi }{4}{{\log }_{e}}2$

Answer 1

Hint: To solve this question, we should know the method of substitutions in integration. In this question, we should substitute ${{x}^{2}}=t\Rightarrow 2xdx=dt$ and after that, we can convert the inverse trigonometric function into an inverse tan function. After that, we can adjust the function so that we can split the function into a sum of two inverse tan functions and integrate it.

  Complete step-by-step answer:
We are given the integral $\int\limits_{0}^{1}{x{{\cot }^{-1}}\left( 1-{{x}^{2}}+{{x}^{4}} \right)dx}$. To solve this integral, we should substitute ${{x}^{2}}=t$. Differentiating it, we get
$\begin{align}
  & 2xdx=dt \\
 & \Rightarrow xdx=\dfrac{dt}{2} \\
\end{align}$.
The limits will be changed as
$\begin{align}
  & x=0 \\
 & \Rightarrow {{x}^{2}}=t=0 \\
\end{align}$
$\begin{align}
  & x=1 \\
 & \Rightarrow {{x}^{2}}=t=1 \\
\end{align}$
Using them in the above integral, we get
$I=\dfrac{1}{2}\int\limits_{0}^{1}{{{\cot }^{-1}}\left( 1-t+{{t}^{2}} \right)dt}$
We know the relation between inverse trigonometric functions of tan and cot.
${{\cot }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{1}{x} \right)$.
This relation is valid for positive values of x. In our question, the integral is from 0 to 1. So, we can apply this condition on the above integral. The integral becomes
$I=\dfrac{1}{2}\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \dfrac{1}{1-t+{{t}^{2}}} \right)dt}$
We can write the numerator as $1=t-\left( t-1 \right)$, we get
$I=\dfrac{1}{2}\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \dfrac{t-\left( t-1 \right)}{1+t\left( t-1 \right)} \right)dt}$
We know the formula related to inverse tan function as
${{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)={{\tan }^{-1}}a-{{\tan }^{-1}}b$ where $ab > -1$
In our question we know that $0 < t < 1$ and $t\left( t-1 \right) > -1$ for all values of t.
Using this relation, we get
$I=\dfrac{1}{2}\int\limits_{0}^{1}{\left( {{\tan }^{-1}}t-{{\tan }^{-1}}\left( t-1 \right) \right)dt}$
To integrate ${{\tan }^{-1}}t$, we should use integration by parts. It is
$\int{uvdx=u\int{vdx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}}}dx$
In our question, we should use $u={{\tan }^{-1}}t,v=1$. We get
$\begin{align}
  & \int{1.{{\tan }^{-1}}t=t{{\tan }^{-1}}t-\int{t\dfrac{1}{1+{{t}^{2}}}dt}}=t{{\tan }^{-1}}t-\dfrac{1}{2}\int{\dfrac{2t}{1+{{t}^{2}}}}dt \\
 & \Rightarrow \int{{{\tan }^{-1}}t=}t{{\tan }^{-1}}t-\dfrac{1}{2}\int{\dfrac{1}{1+{{t}^{2}}}d\left( 1+{{t}^{2}} \right)=}t{{\tan }^{-1}}t-\dfrac{1}{2}\ln \left( 1+{{t}^{2}} \right) \\
\end{align}$
Using this result in the above integral, we get
$\begin{align}
  & I=\dfrac{1}{2}\int\limits_{0}^{1}{\left( {{\tan }^{-1}}t-{{\tan }^{-1}}\left( t-1 \right) \right)dt} \\
 & \Rightarrow I=\dfrac{1}{2}\left[ t{{\tan }^{-1}}t-\dfrac{1}{2}\ln \left( 1+{{t}^{2}} \right) \right]_{0}^{1}-\dfrac{1}{2}\int\limits_{0}^{1}{{{\tan }^{-1}}\left( t-1 \right)}dt \\
\end{align}$
We can use the formula $\int{{{\tan }^{-1}}t=}t{{\tan }^{-1}}t-\dfrac{1}{2}\int{\dfrac{1}{1+{{t}^{2}}}d\left( 1+{{t}^{2}} \right)=}t{{\tan }^{-1}}t-\dfrac{1}{2}\ln \left( 1+{{t}^{2}} \right)$ by substituting $t=t-1$ as the coefficients and the powers of t are same.
The integral becomes
$\begin{align}
  & I=\dfrac{1}{2}\left[ 1{{\tan }^{-1}}1-\dfrac{1}{2}\ln \left( 1+{{1}^{2}} \right)-\left( 0{{\tan }^{-1}}0-\dfrac{1}{2}\ln \left( 1+{{0}^{2}} \right) \right) \right]_{0}^{1}-\dfrac{1}{2}\left[ \left( t-1 \right){{\tan }^{-1}}\left( t-1 \right)-\dfrac{1}{2}\ln \left( 1+{{\left( t-1 \right)}^{2}} \right) \right]_{0}^{1} \\
 & \Rightarrow I=\dfrac{1}{2}\left[ \dfrac{\pi }{4}-\dfrac{1}{2}\ln \left( 2 \right) \right]-\dfrac{1}{2}\left[ \left( 0{{\tan }^{-1}}0-\dfrac{1}{2}\ln \left( 1+{{0}^{2}} \right) \right)-\left( -1{{\tan }^{-1}}\left( -1 \right)-\dfrac{1}{2}\ln \left( 1+{{1}^{2}} \right) \right) \right] \\
 & \Rightarrow I=\dfrac{1}{2}\left[ \dfrac{\pi }{4}-\dfrac{1}{2}\ln \left( 2 \right) \right]-\dfrac{1}{2}\left[ -\left( \dfrac{\pi }{4}-\dfrac{1}{2}\ln \left( 2 \right) \right) \right] \\
 & \Rightarrow I=\dfrac{1}{2}\left[ \dfrac{\pi }{4}-\dfrac{1}{2}\ln \left( 2 \right) \right]+\dfrac{1}{2}\left[ \dfrac{\pi }{4}-\dfrac{1}{2}\ln \left( 2 \right) \right]=\dfrac{\pi }{4}-\dfrac{1}{2}\ln \left( 2 \right) \\
\end{align}$

$\therefore$ The value of the integral is $\dfrac{\pi }{4}-\dfrac{1}{2}\ln \left( 2 \right)$.

  So, the correct answer is “Option A”.

  Note: Some students can make mistakes while substituting the limits in the final integration. If we ignore the negative sign in the above formula, we might end up getting a value of zero. So, we should do step by step while substituting the limits. While doing the integration of ${{\tan }^{-1}}t$, we should know that the selection of u and v functions play a major role in solving the integral. If we exchange the values of u and v, we will not get the answer. So, there is a basic rule in integration by parts which is known as ILATE rule. It means that the order of preference for selecting a function as u in integration by parts is Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential function. We should use this preference order while selecting the function as u.