Question

Evaluate the following integral
\[\int\limits_{0}^{1}{x{{\tan }^{-1}}xdx=}\]
(a) \[\dfrac{\pi }{4}+\dfrac{1}{2}\]
(b) \[\dfrac{\pi }{4}-\dfrac{1}{2}\]
(c) \[\dfrac{1}{2}-\dfrac{\pi }{4}\]
(d) \[-\dfrac{\pi }{4}-+\dfrac{1}{2}\]

Answer 1

Hint: In this question we can see that there are 2 functions x and \[{{\tan }^{-1}}x\]. Assume x = v and \[{{\tan }^{-1}}x=u\] and use the formula of integration by parts, given by \[\int{uv\text{ }dx}=u\int{v\text{ }dx}-\int{\left[ \dfrac{du}{dx}\int{v\text{ }dx} \right]}\text{ }dx\]. Substitute the values of u and v and solve accordingly.

Complete step-by-step answer:

Here, we have to solve the integral \[\int\limits_{0}^{1}{x{{\tan }^{-1}}xdx}\]. Let us consider the integral given in the question as \[I=\int\limits_{0}^{1}{x{{\tan }^{-1}}x}\text{ }dx\].
Whenever we have the product of two functions in the integral, we can use the integration by parts method to evaluate it. While choosing the functions, we must follow the ILATE rule. So, I stand for Inverse function, L for Logarithmic function, A for Algebraic function, T for Trigonometric function and E for exponential function. It tells us the order in which the first function is to be chosen, the first being I and then last being E. So, here we will consider the first function as \[{{\tan }^{-1}}x\] as it is an inverse function.
We know that integration by parts is \[\int{u.v\text{ }dx}=u\int{v\text{ }dx}-\int{\left[ \dfrac{du}{dx}.\int{v\text{ }dx} \right]}\text{ }dx\]. We can apply this by taking \[u={{\tan }^{-1}}x\] and v = x,
\[I={{\tan }^{-1}}x\int{x}\text{ }dx-\int{\left[ \dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}.\int{x}\text{ }dx \right]}\text{ }dx\]
We know that \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\] and \[\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}\]. By applying this in the above equation, we get,
\[I=\left( {{\tan }^{-1}}x \right)\left( \dfrac{{{x}^{2}}}{2} \right)-\int{\left[ \left( \dfrac{1}{1+{{x}^{2}}} \right)\left( \dfrac{{{x}^{2}}}{2} \right) \right]dx}\]
\[I=\left( {{\tan }^{-1}}x \right)\left( \dfrac{{{x}^{2}}}{2} \right)-\int{\dfrac{{{x}^{2}}}{2\left( 1+{{x}^{2}} \right)}dx}\]
Let us take \[\int{\dfrac{{{x}^{2}}}{2\left( 1+{{x}^{2}} \right)}dx=G}\]
So we get, \[I=\left( {{\tan }^{-1}}x \right)\left( \dfrac{{{x}^{2}}}{2} \right)-G....\left( i \right)\]
By adding and subtracting 1 in the numerator and in LHS of G, we get \[G=\int{\dfrac{\left( {{x}^{2}}+1 \right)-\left( 1 \right)}{2\left( 1+{{x}^{2}} \right)}}dx\]. By separating RHS intro fractions, we get,
\[G=\dfrac{1}{2}\int{\left[ \dfrac{\left( {{x}^{2}}+1 \right)}{\left( {{x}^{2}}+1 \right)}-\dfrac{1}{\left( {{x}^{2}}+1 \right)} \right]}\text{ }dx\]
\[G=\dfrac{1}{2}\left[ \int{1\text{ }dx}-\int{\dfrac{1}{\left( 1+{{x}^{2}} \right)}}dx \right]\]
We know that \[\int{k}\text{ }dx=kx\] and \[\int{\dfrac{1}{\left( 1+{{x}^{2}} \right)}dx={{\tan }^{-1}}x}\]. By applying this is the above equation we get, \[G=\dfrac{1}{2}\left[ x-{{\tan }^{-1}}x \right]\]
By substituting the value of G in equation (i), we get,
\[I=\left( {{\tan }^{-1}}x \right)\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{1}{2}\left( x-{{\tan }^{-1}}x \right)\]
\[I=\dfrac{1}{2}\left( {{x}^{2}}{{\tan }^{-1}}x-x+{{\tan }^{-1}}x \right)\]
Now by applying limits from 0 to 1 in the above integral, we get,
\[I=\dfrac{1}{2}\left[ \left( {{x}^{2}}{{\tan }^{-1}}x-x+{{\tan }^{-1}}x \right) \right]_{0}^{1}\]
\[I=\dfrac{1}{2}\left[ 1.\left( {{\tan }^{-1}}\left( 1 \right) \right)-1+{{\tan }^{-1}}\left( 1 \right) \right]-\left[ 0.{{\tan }^{-1}}0-0+{{\tan }^{-1}}\left( 0 \right) \right]\]
We know that \[{{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}\]. By applying this in the above equation, we get,
\[I=\dfrac{1}{2}\left[ \left( \dfrac{\pi }{4}-1+\dfrac{\pi }{4} \right)-\left( 0 \right) \right]\]
\[\Rightarrow I=\dfrac{1}{2}\left[ \dfrac{2\pi }{4}-1 \right]\]
\[\Rightarrow I=\dfrac{1}{2}\left[ \dfrac{\pi }{2}-1 \right]\]
Again simplifying we get \[I=\dfrac{\pi }{4}-\dfrac{1}{2}\]. Hence, we get the value of our integral as \[\left( \dfrac{\pi }{4}-\dfrac{1}{2} \right)\].
Therefore, option (b) is the right answer.

Note: Students are advised to apply the limits in the last step of integration to avoid confusion. Also while using integration by parts, choose u and v such that \[\int{v\text{ }dx}\] and \[\int{\left[ \dfrac{du}{dx}\int{v\text{ }dx} \right]dx}\] is simple to integrate. This is generally obtained by keeping the order of u and v as per the ILATE rule. Any silly mistake while applying the limits should be avoided.