Answer
398.4k+ 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
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science 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)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 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)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![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)
How do you graph the function fx 4x class 9 maths 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)