Question

Find the value of $A$ in this derivative
$\int\limits_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{x^4}}}{{1 - {x^4}}}} {\cos ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}}dx = \dfrac{A}{{\sqrt 3 }} + \dfrac{{{A^2}}}{{12}} - \dfrac{A}{4}\log \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$
A. $\pi {\text{ }}$
B. $2\pi$
C. $3\pi$
D. ${\text{None of these}}$

Answer 1

we have to find the value of A in this sum we need to solve using both side similarities. Before that we have to know about integration, Integration is a derivative which increases the value and it processes on with the limits. First we need to know about odd and even functions. To find out whether the function is odd or even the interval must be in the form of $\left[ { - a,a} \right]$. If for that there will be formulas to solve and limits which should be solved in the function by complete step-by-step solution.

Formula used: ${\cos ^{ - 1}}y = \dfrac{\pi }{2}{\sin ^{ - 1}}y$
${\tan ^{ - 1}}x = \dfrac{x}{{1 + {x^2}}}$
${a^2} - {b^2} = (a + b)(a - b)$
${\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}$
$\int {\dfrac{{dx}}{{{a^2} - {x^2}}} = \dfrac{1}{{2a}}\log \dfrac{{a + x}}{{a - x}}}$

Complete step-by-step answer:
We have to find the value of $A$ in the given derivative. The given derivative is $\int\limits_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{x^4}}}{{1 - {x^4}}}} {\cos ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}}dx = \dfrac{A}{{\sqrt 3 }} + \dfrac{{{A^2}}}{{12}} - \dfrac{A}{4}\log \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$
First we need to convert the derivation from ${\cos ^{ - 1}}$ to ${\sin ^{ - 1}}$, because for this type of sum we need think reverse, LHS
$\Rightarrow \int\limits_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{{x^4}}}{{1 - {x^4}}}} {\cos ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}}dx$
We have solve separately,
Let us consider, ${\cos ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}}$
Now use the formula mentioned in formulas used
$\Rightarrow {\cos ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}} = \dfrac{\pi }{2} - {\sin ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}}$
Now, again use ${\tan ^{ - 1}}x$ formula mention in formula used
Then, ${\sin ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}} = 2{\tan ^{ - 1}}x$
Then, the derivation becomes
$\Rightarrow \int\limits_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\left[ {\dfrac{{{x^4}}}{{1 - {x^4}}}\left( {\dfrac{\pi }{2} - 2{{\tan }^{ - 1}}x} \right)} \right]} {\text{ }}$
Now multiply the fraction in the bracket variable, we get
$\Rightarrow \int\limits_{ - \dfrac{1}{{\sqrt 3 }}}^{\dfrac{1}{{\sqrt 3 }}} {\left[ {\dfrac{\pi }{2}\dfrac{{{x^4}}}{{1 - {x^4}}} - \dfrac{{{x^4}}}{{1 - {x^4}}}2{{\tan }^{ - 1}}x} \right]} {\text{ }}dx$
Now, applying odd and even function, that is
If ${\text{ }}f( - x) = f(x)$, the function is even
If ${\text{ }}f( - x) = - f(x)$, the function is odd
By, applying this concept, we can able to check the function is odd or even
If a function becomes odd, then the function will becomes zero,
Then the function will becomes twice the value,
That is, $f(x) = 0$
$\because \dfrac{{{x^4}}}{{1 - x}}2{\tan ^{ - 1}}x$ is an odd function.
Then, $\therefore \dfrac{{{x^4}}}{{1 - x}}2{\tan ^{ - 1}}x{\text{ = 0}}$
If a integral function is even, then $\int\limits_{ - a}^a {f(x) = 2\int\limits_0^a {f(x)} }$
Then, then function $\dfrac{\pi }{2}\dfrac{{{x^4}}}{{1 - {x^4}}}{(h - 3)^2} + {(k - 4)^2} = 50$ becomes even, then we applying integral function on even concept, and taking $\dfrac{\pi }{2}$ out of integral, we get
$\Rightarrow 2.\dfrac{\pi }{2}\int\limits_0^{\dfrac{1}{{\sqrt 3 }}} {\left( { - 1 + \dfrac{1}{{1 - {x^4}}}} \right){\text{ }}} dx + 0$
Now, multiplying 2 inside the function, we get
$\Rightarrow \dfrac{\pi }{2}\int\limits_0^{\dfrac{1}{{\sqrt 3 }}} {\left( { - 2 + \dfrac{2}{{1 - {x^4}}}} \right)} \;dx + 0$
Now, consider
$\Rightarrow \dfrac{2}{{1 - {x^4}}} = \dfrac{2}{{(1 - {x^2})(1 + {x^2})}}$
Once again, use the formula mentioned in formula used, then we need add and sub ${x^2}$ arbitrary, which means imaginary
$\Rightarrow \dfrac{2}{{1 - {x^4}}} = \dfrac{{1 + 1 + {x^2} - {x^2}}}{{(1 - {x^2})(1 + {x^2})}}$
$\Rightarrow \dfrac{2}{{1 - {x^4}}} = \dfrac{1}{{1 - {x^2}}} + \dfrac{1}{{1 + {x^2}}}$
Now substitute the value in the derivation, we get
$\Rightarrow \dfrac{\pi }{2}\int\limits_0^{\dfrac{1}{{\sqrt 3 }}} {\left( { - 2 + \dfrac{1}{{1 - {x^2}}} + \dfrac{1}{{1 + {x^2}}}} \right)\;} dx$
Now, once use formula mentioned in formula used and then integrate, we get
$\Rightarrow \dfrac{\pi }{2}\left[ { - 2x + \dfrac{1}{{2.1}}\log \dfrac{{1 + x}}{{1 - x}} + {{\tan }^{ - 1}}x} \right]_0^{\dfrac{1}{{\sqrt 3 }}}$
Now substituting limit values, for substituting limit values for $0$ then the whole values becomes zero, then only$\dfrac{1}{{\sqrt 3 }}$values get substitute in the derivation for variable $x$,
$\Rightarrow \dfrac{\pi }{2}\left[ {\dfrac{{ - 2}}{{\sqrt 3 }} + \dfrac{1}{2}\log \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}} + \dfrac{\pi }{6}} \right]$
Now, multiply $\dfrac{\pi }{2}$ in the bracket expression after rearranging, we get
$\Rightarrow \dfrac{{{\pi ^2}}}{{12}} + \dfrac{\pi }{3} - \dfrac{\pi }{2}\log \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$
Now, look at the question,
RHS, Compare with the solution, then the value of $A$ is $\pi$
Thus the correct answer for a value $A$ is $\pi$

So, the correct answer is “Option A”.

Note: The solution needs to be determined in the manner of the reverse method and also these kinds of problems need attention step by step. The problem has to be simply on an idea based on another side it may lead may step and change in step by according to the question we need to find the missing value. So that the problem on integral function needs to be more about ideas on formulas that must be relocated according to the problem.