The value of $\int\limits_0^1 {\left[ {\sqrt {\dfrac{{1 - x}}{{1 + x}}} } \right]} dx$is
1. $\left( {\dfrac{\pi }{2}} \right) + 1$
2. $\left( {\dfrac{\pi }{2}} \right) - 1$
3. $ - 1$
4. $1$

Hint: Assume the given integral is equal to $I$. Then, to solve the integral let $x = \cos 2\theta $ and find the first derivative to put the values in given integral. After putting the values convert the limits and apply trigonometric identities $\cos 2A = 2{\cos ^2}A - 1$, $\cos 2A = 1 - 2{\sin ^2}A$ to solve further.

Formula Used:
Trigonometric formula –
$\cos 2A = 2{\cos ^2}A - 1$
$\cos 2A = 1 - 2{\sin ^2}A$
$\sin 2A = 2\sin A\cos A$
Differentiation formula –
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
Integration formula –
$\int {f\left( {g\left( x \right)} \right)dx = \dfrac{{\int {f\left( {g\left( x \right)} \right)dx} }}{{\dfrac{d}{{dx}}g\left( x \right)}}} $

Complete step by step Solution:
Given that,
$I = \int\limits_0^1 {\left[ {\sqrt {\dfrac{{1 - x}}{{1 + x}}} } \right]} dx - - - - - \left( 1 \right)$
Let, $x = \cos 2\theta - - - - - \left( 2 \right)$
Differentiate equation (2) with respect to $x$,
$1 = - \sin 2\theta \times \left( {2\dfrac{{d\theta }}{{dx}}} \right)$
$dx = - 2\sin 2\theta d\theta $
Now the Limits will also change,
At $x = 0$
$0 = \cos 2\theta $
$\cos \dfrac{\pi }{2} = \cos 2\theta $
And at $\theta = \dfrac{\pi }{4}$
$x = 1$
$1 = \cos 2\theta $
$\cos {0^ \circ } = \cos 2\theta $
$\theta = 0$
Equation (1) will be
$I = \int\limits_{\dfrac{\pi }{4}}^0 {\sqrt {\dfrac{{1 - \cos 2\theta }}{{1 + \cos 2\theta }}} \left( { - 2\sin 2\theta } \right)} d\theta $
Applying $1 - \cos 2\theta = 2{\sin ^2}\theta $, $1 + \cos 2\theta = 2{\cos ^2}\theta $ in the above integral
$ = \int\limits_{\dfrac{\pi }{4}}^0 {\sqrt {\dfrac{{2{{\sin }^2}\theta }}{{2{{\cos }^2}\theta }}} \left( { - 2\sin 2\theta } \right)} d\theta $
Now, apply $\sin 2\theta = 2\sin \theta \cos \theta $
$ = \int\limits_{\dfrac{\pi }{4}}^0 {\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)\left( { - 2\sin 2\theta } \right)} d\theta $
$ = - 2\int\limits_{\dfrac{\pi }{4}}^0 {\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)\left( {2\sin \theta \cos \theta } \right)} d\theta $
$ = - 2\int\limits_{\dfrac{\pi }{4}}^0 {2{{\sin }^2}\theta } d\theta $
$ = - 2\int\limits_{\dfrac{\pi }{4}}^0 {\left( {1 - \cos 2\theta } \right)} d\theta $
On integrating, now we get
$ = - 2\left[ {\theta - \dfrac{{\sin 2\theta }}{2}} \right]_{\dfrac{\pi }{4}}^0$
Put the limits in required integration and use the format of upper limit $ - $lower limit
$ = - 2\left[ {0 - \left( {\dfrac{\pi }{4} - \dfrac{1}{2}} \right)} \right]$
$ = \dfrac{\pi }{2} - 1$

Hence, the correct option is 2.

Note: In such a question, students must have a good knowledge of trigonometric formulas and should also know that the integration of $\sin x$is $ - \cos x$ and $\cos x$ is $\sin x$. Also, if you are taking the value of a given variable as any function don’t forget to convert the limits. Otherwise, the solution will be incorrect. Then enter the integration limit and solve algebraically to get the answer to the question.