Answer
385.5k+ views
Hint: 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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)