Question

The value of $\int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{x^4}}}{{1 - {x^4}}}} {\cos ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}}dx = \dfrac{A}{{\sqrt 3 }} + \dfrac{{{A^2}}}{{12}} - \dfrac{A}{4}\log \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$, then the value of A is:
A. $\pi $
B. $2\pi $
C. $3\pi $
D. None of these

Answer 1

Hint: We will first make the modification of putting x = -y and then put in the values accordingly, we will somehow get two times the integral we require to a definite value whose integral we can find easily and then compare that to the given RHS and find A.

Complete step-by-step answer:
Let $I = \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{x^4}}}{{1 - {x^4}}}} {\cos ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}}dx$ ……….(1)
Let us now put $x = - y$, then we will have to put $dx = - dy$ and the limits will interchange as well because when x will go to a, then –y will go to –a.
Hence, we will have:- $I = \int_{\dfrac{1}{{\sqrt 3 }}}^{ - \dfrac{1}{{\sqrt 3 }}} {\dfrac{{{y^4}}}{{1 - {y^4}}}} {\cos ^{ - 1}}\dfrac{{ - 2y}}{{1 + {y^2}}}( - dy)$
We know that ${\cos ^{ - 1}}( - x) = \pi - {\cos ^{ - 1}}x$ for $ - 1 \leqslant x \leqslant 1$.
Using this, we will get:-
$I = \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{y^4}}}{{1 - {y^4}}}} \left[ {\pi - {{\cos }^{ - 1}}\left( {\dfrac{{2y}}{{1 + {y^2}}}} \right)} \right]dy$
Opening up the bracket on RHS will lead us to:-
$I = \pi \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{y^4}}}{{1 - {y^4}}}} dy - \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{y^4}}}{{1 - {y^4}}}{{\cos }^{ - 1}}\left( {\dfrac{{2y}}{{1 + {y^2}}}} \right)} dy$ …………(2)
Let ${I_1} = \pi \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{y^4}}}{{1 - {y^4}}}} dy$ and ${I_2} = \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{y^4}}}{{1 - {y^4}}}{{\cos }^{ - 1}}\left( {\dfrac{{2y}}{{1 + {y^2}}}} \right)} dy$
Replacing y by x in ${I_1}$ and ${I_2}$ will not change anything and we will get:-
${I_1} = \pi \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{x^4}}}{{1 - {x^4}}}} dx$ ………….(3)
${I_2} = \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{x^4}}}{{1 - {x^4}}}{{\cos }^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)} dx$ ……….(4)
Now, let we can clearly see that ${I_2} = I$ ……………(5)
Putting (3), (4) and (5) in (2), we will get:-
$I = \pi \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{x^4}}}{{1 - {x^4}}}} dx - I$
Taking the I from RHS to LHS, we will get:-
$ \Rightarrow 2I = \pi \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{x^4}}}{{1 - {x^4}}}} dx$
We can rewrite this as:-
$ \Rightarrow 2I = \pi \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{1 + {x^4} - 1}}{{1 - {x^4}}}} dx$
Now, rewriting it as:-
$ \Rightarrow 2I = \pi \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\left[ { - 1 + \dfrac{1}{{1 - {x^4}}}} \right]} dx$
Now, rewriting it again as follows:-
$ \Rightarrow 2I = - \pi \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {dx + \pi \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{1}{{1 - {x^4}}}} dx} $
On simplifying the RHS, we will get:-
$ \Rightarrow 2I = - \dfrac{2}{{\sqrt 3 }}\pi + \pi \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{1}{{1 - {x^4}}}} dx$ …………(6)
Let ${I_3} = \int_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{1}{{1 - {x^4}}}} dx$
We can see that ${I_3}$ is an even function because putting –x does not change it at all and we also know that: $\int_{ - a}^a {f(x)dx = 2} \int_0^a {f(x)dx} $ if $f(x)$ is an even function.
Therefore, we have:-
${I_3} = 2\int_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{1}{{1 - {x^4}}}} dx$
We can rewrite it as:-
$ \Rightarrow {I_3} = \int_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{1 + 1 + {x^2} - {x^2}}}{{1 - {x^4}}}} dx$
Now we will use the formula:- ${a^2} - {b^2} = (a + b)(a - b)$
Therefore, we get:- $ \Rightarrow {I_3} = \int_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{1 + 1 + {x^2} - {x^2}}}{{(1 - {x^2})(1 + {x^2})}}} dx$
Rewriting it as follows:-
$ \Rightarrow {I_3} = \int_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{1}{{1 - {x^2}}}} dx + \int_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{1}{{1 + {x^2}}}} dx$ …………..(7)
Now, we will use the formula that:- $\int {\dfrac{1}{{1 + {x^2}}}dx} = {\tan ^{ - 1}}x$
Putting this in (7), we will get that:-
$ \Rightarrow {I_3} = \int_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{1}{{1 - {x^2}}}} dx + {\tan ^{ - 1}}\dfrac{1}{{\sqrt 3 }} - {\tan ^{ - 1}}0$
On simplifying it, we will get:-
$ \Rightarrow {I_3} = \int_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{1}{{1 - {x^2}}}} dx + \dfrac{\pi }{6}$ ……….(8)
Now, let ${I_4} = \int_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{1}{{1 - {x^2}}}} dx$
We know that $\dfrac{1}{{1 - {x^2}}} = \dfrac{E}{{1 - x}} + \dfrac{F}{{1 + x}}$
On solving this, we will have:-
$1 = E(1 + x) + F(1 - x)$
Comparing both and solving, we will get:-
$\dfrac{1}{{1 - {x^2}}} = \dfrac{1}{{2(1 - x)}} + \dfrac{1}{{2(1 + x)}}$
Putting this in (8), we will get:-
${I_4} = \int_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{1}{{2(1 - x)}}} dx + \int_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{1}{{2(1 + x)}}} dx$
Now, we know that $\int {\dfrac{1}{x}dx} = \ln x$
So, we will get:- $\int {\dfrac{1}{{2(1 - x)}}dx} + \int {\dfrac{1}{{2(1 + x)}}dx} = - \dfrac{1}{2}\ln \left| {1 - x} \right| + \dfrac{1}{2}\ln \left| {1 + x} \right|$
Hence, $\int {\dfrac{1}{{2(1 - x)}}dx} + \int {\dfrac{1}{{2(1 + x)}}dx} = \dfrac{1}{2}\dfrac{{\ln \left| {1 + x} \right|}}{{\ln \left| {1 - x} \right|}}$ ( Because $\ln a - \ln b = \ln \dfrac{a}{b}$ )
So, ${I_4} = \dfrac{1}{2}\dfrac{{\ln \left| {\sqrt 3 + 1} \right|}}{{\ln \left| {\sqrt 3 - 1} \right|}}$
On rewriting it, we will get:-
${I_4} = - \dfrac{1}{2}\dfrac{{\ln \left| {\sqrt 3 - 1} \right|}}{{\ln \left| {\sqrt 3 + 1} \right|}}$ (Because $\ln {a^{ - 1}} = - \ln a$)
Putting this in (8), we will get:-
$ \Rightarrow {I_3} = - \dfrac{1}{2}\dfrac{{\ln \left| {\sqrt 3 - 1} \right|}}{{\ln \left| {\sqrt 3 + 1} \right|}} + \dfrac{\pi }{6}$
Now putting this in (6), we will get:-
$ \Rightarrow 2I = - \dfrac{2}{{\sqrt 3 }}\pi - \dfrac{\pi }{2}\dfrac{{\ln \left| {\sqrt 3 - 1} \right|}}{{\ln \left| {\sqrt 3 + 1} \right|}} + \dfrac{{{\pi ^2}}}{6}$
Hence, we get:-
$ \Rightarrow I = - \dfrac{1}{{\sqrt 3 }}\pi - \dfrac{\pi }{4}\dfrac{{\ln \left| {\sqrt 3 - 1} \right|}}{{\ln \left| {\sqrt 3 + 1} \right|}} + \dfrac{{{\pi ^2}}}{{12}}$
Now comparing this to $\dfrac{A}{{\sqrt 3 }} + \dfrac{{{A^2}}}{{12}} - \dfrac{A}{4}\log \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$, we can clearly see that $A = \pi $.

So, the correct answer is “Option A”.

Note: The students must note that we have created a lot of I’s to reach the required integral, they may go in with the single integral as well but that can create a lot of confusion.
Alternate Way:-
Let us talk about the alternate method in brief:
We will use that:- ${\cos ^{ - 1}}y = \dfrac{\pi }{2} - {\sin ^{ - 1}}y$.
So, ${\cos ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) = \dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) = \dfrac{\pi }{2} - {\tan ^{ - 1}}x$ (By formula)
Now, use the fact that $\dfrac{{{x^4}}}{{1 - {x^4}}} 2{\tan ^{ - 1}}x$ is an odd function and keep on simplifying, you will get the required answer.