Question

Find $\int\limits_0^{\dfrac{1}{2}} {\dfrac{{dx}}{{(1 + {x^2})\sqrt {1 - {x^2}} }}}$ is equal to

1. $\dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\sqrt {\dfrac{2}{3}} } \right) + C$
2. $\dfrac{2}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{3}{{\sqrt 2 }}} \right) + C$
3. $\dfrac{2}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{3}{2}} \right) + C$
4. $\dfrac{2}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) + C$

Answer 1

In the above problem substitution method is used to reduce the given definite integration.
In place of the variable x we will substitute the value of $\sin \theta$ to reduce the given integral in the simplified form, which is called the substitution method.
Using the substitution method we will solve the given problem.

Complete step by step answer:
Let's do the substitution of the x variable and then we will solve the problem.
$\Rightarrow \int\limits_0^{\dfrac{1}{2}} {\dfrac{{dx}}{{(1 + {x^2})\sqrt {1 - {x^2}} }}}$(given integral)
Let's keep $x = \sin \theta$ , therefore limits of the integration will also going to be changed,
$\Rightarrow {\sin ^{ - 1}}x = \theta$ , when x = 0 ,$\theta$ also equals to zero, x =1/2, $\theta$=${30^0}$
$\Rightarrow \int\limits_0^{\dfrac{\pi }{6}} {\dfrac{{\cos \theta d\theta }}{{(1 + {{\sin }^2}\theta )\sqrt {(1 - {{\sin }^2}\theta )} }}}$ (we will cancel both the cosine terms of the numerator and denominator)
$\Rightarrow \int\limits_0^{\dfrac{\pi }{6}} {\dfrac{{\cos \theta d\theta }}{{(1 + {{\sin }^2}\theta )\cos \theta }}}$ (${\cos ^2}\theta = 1 - {\sin ^2}\theta$)
$\Rightarrow \int\limits_0^{\dfrac{\pi }{6}} {\dfrac{{d\theta }}{{(1 + {{\sin }^2}\theta )}}}$(we will multiply both numerator and denominator by ${\sec ^2}\theta$ )
$\Rightarrow \int\limits_0^{\dfrac{\pi }{6}} {\dfrac{{{{\sec }^2}\theta d\theta }}{{({{\sec }^2}\theta + {{\sec }^2}\theta {{\sin }^2}\theta )}}}$
$\Rightarrow \int\limits_0^{\dfrac{\pi }{6}} {\dfrac{{{{\sec }^2}\theta d\theta }}{{({{\sec }^2}\theta + {{\tan }^2}\theta )}}} ({\sec ^2}\theta \; is \;the \;reciprocal \;of \;{\cos ^2}\theta )$
$\Rightarrow \int\limits_0^{\dfrac{\pi }{6}} {\dfrac{{{{\sec }^2}\theta d\theta }}{{(1 + 2{{\tan }^2}\theta )}}}$(${\sec ^2}\theta$=$1 + {\tan ^2}\theta$ )
Now, we replace tan$\theta$ as u,
$\Rightarrow \tan \theta = u,\theta = 0$
$\Rightarrow u = 0$
When $\tan \dfrac{\pi }{6},u = \dfrac{1}{{\sqrt 3 }}$
$\Rightarrow \int\limits_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{du}}{{(1 + 2{u^2})}}}$ (we will take 2 common out of the denominator)
$\Rightarrow \int\limits_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{du}}{{2(\dfrac{1}{2} + {u^2})}}}$(We will apply the direct formula of $\dfrac{1}{a}{\tan ^{ - 1}}\dfrac{u}{a}$ , a is the constant here)
$\Rightarrow \int\limits_0^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{du}}{{2({{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2} + {u^2})}}}$
$\Rightarrow [\dfrac{1}{2} \times \dfrac{1}{{\dfrac{1}{{\sqrt 2 }}}}{\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{{\sqrt 3 }}}}{{\dfrac{1}{{\sqrt 2 }}}}} \right) - \dfrac{1}{2} \times \dfrac{1}{{\dfrac{1}{{\sqrt 2 }}}}{\tan ^{ - 1}}\left( {\dfrac{0}{{\dfrac{1}{{\sqrt 2 }}}}} \right)]$

We have substituted the limits in the formula of integration, we obtain
$\Rightarrow \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\sqrt {\dfrac{2}{3}} } \right) + C$

So, the correct answer is Option 1.

Note: Integration has another method to solve the problem when two functions are simultaneously given which is called Integration by parts. It has many other applications in problem solving such as in deriving the Euler-Lagrange equation, for determining the boundary conditions in Sturm-Liouville theory, it is used in operator theory, decay of Fourier transform, Fourier transform of derivative, used in harmonics analysis, to find the gamma function identity etc.