Question

Solve the following definite integral:
\[\int\limits_0^{\dfrac{1}{2}} {\dfrac{{dx}}{{(1 + {x^2})\sqrt {1 - {x^2}} }}} \]
A) \[\dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\sqrt {\dfrac{2}{3}} \]
B)\[\dfrac{2}{{\sqrt 2 }}{\tan ^{ - 1}}(\dfrac{3}{{\sqrt 2 }})\]
C) \[\dfrac{{\sqrt 2 }}{2}{\tan ^{ - 1}}(\dfrac{3}{2})\]
D) \[\dfrac{{\sqrt 2 }}{2}{\tan ^{ - 1}}(\dfrac{{\sqrt 3 }}{2})\]

Answer 1

Hint: To solve this integral, we have to substitute the value of x=\[\dfrac{1}{t}\] and further, we will apply the integration formulae given as below-
 \[\int\limits_0^x {\dfrac{{dx}}{{{x^2} + {a^2}}}} \]=\[\dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a}\]

Complete step by step solution: Given definite integral in the question,
I=\[\int\limits_0^{\dfrac{1}{2}} {\dfrac{{dx}}{{(1 + {x^2})\sqrt {1 - {x^2}} }}} \]
Putting the value of x = \[\dfrac{1}{t}\] and differentiating it, dx = \[\dfrac{{ - dt}}{{{t^2}}}\], we will get
\[ \Rightarrow \int\limits_0^{\dfrac{1}{2}} {\dfrac{{\dfrac{{ - dt}}{{{t^2}}}}}{{(1 + \dfrac{1}{{{t^2}}})\sqrt {1 - \dfrac{1}{{{t^2}}}} }}} \]
 Now, taking l.c.m in denominator and solving it, we will get,
 \[ \Rightarrow \int\limits_0^{\dfrac{1}{2}} {\dfrac{{\dfrac{{ - dt}}{{{t^2}}}}}{{(\dfrac{{{t^2} + 1}}{{{t^2}}})\sqrt {\dfrac{{{t^2} - 1}}{{{t^2}}}} }}} \] \[\]
hence, on further solving it, it will be deduced to \[\int\limits_0^{\dfrac{1}{2}} {\dfrac{{ - tdt}}{{(1 + {t^2})\sqrt {{t^2} - 1} }}} \]
In this case, we need to have further substitution by substituting \[{t^2} - 1 = {u^2}\]
\[ \Rightarrow {t^2} = {u^2} + 1\]
Differentiating it we will find, \[tdt = udu\]
Now, substituting all these values, the integral will become

\[ \Rightarrow \int\limits_0^{\dfrac{1}{2}} {\dfrac{{ - udu}}{{(1 + {u^2} + 1)\sqrt {{u^2}} }}} \]
As, we will solve the integral in terms of u, we will determine \[\int\limits_0^{\dfrac{1}{2}} {\dfrac{{ - du}}{{{u^2} + 2}}} \]
 Now, we will use the integration formulae which can be stated as
\[\int\limits_0^x {\dfrac{{dx}}{{{x^2} + {a^2}}}} \]=\[\dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a}\]

So, our integral will be in terms of u will become,
I = \[\left. {\dfrac{1}{{\sqrt 2 }}{{\tan }^{ - 1}}\dfrac{{( - u)}}{{\sqrt 2 }}} \right|_0^{\dfrac{1}{2}}\]
\[ \Rightarrow \left. {\dfrac{{ - 1}}{{\sqrt 2 }}{{\tan }^{ - 1}}\dfrac{{\sqrt {{t^2} - 1} }}{{\sqrt 2 }}} \right|_0^{\dfrac{1}{2}}\] [on putting the value of u ],
Putting \[{t^2} = \dfrac{1}{{{x^2}}}\]and we will get ,
\[\left. { \Rightarrow \dfrac{{ - 1}}{{\sqrt 2 }}{{\tan }^{ - 1}}\dfrac{{\sqrt {1 - {x^2}} }}{{x\sqrt 2 }}} \right|_0^{\dfrac{1}{2}}\]
Applying both upper and lower limits and due to (-) term, we will get
I = \[\dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\sqrt {\dfrac{2}{3}} \]

So, the correct option is (A).

Note: 1) To solve definite integrals of the form \[\dfrac{1}{{({x^2} + {a^2})(\sqrt {{x^2} - {a^2}} )}}\],we have to suppose the square root term be another variable and then we will further solve it.
2) It can be solved by another type of substitution i.e, x= tanα
     \[\dfrac{1}{{({x^2} + {a^2})(\sqrt {{x^2} - {a^2}} )}}\]