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Question: Solve the following definite integral: \[\int\limits_0^{\dfrac{1}{2}} {\dfrac{{dx}}{{(1 + {x^2})\s...

Solve the following definite integral:
∫012dx(1+x2)1−x2\int\limits_0^{\dfrac{1}{2}} {\dfrac{{dx}}{{(1 + {x^2})\sqrt {1 - {x^2}} }}}
A) 12tan⁡−123\dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\sqrt {\dfrac{2}{3}}
B)22tan⁡−1(32)\dfrac{2}{{\sqrt 2 }}{\tan ^{ - 1}}(\dfrac{3}{{\sqrt 2 }})
C) 22tan⁡−1(32)\dfrac{{\sqrt 2 }}{2}{\tan ^{ - 1}}(\dfrac{3}{2})
D) 22tan⁡−1(32)\dfrac{{\sqrt 2 }}{2}{\tan ^{ - 1}}(\dfrac{{\sqrt 3 }}{2})

Explanation

Solution

To solve this integral, we have to substitute the value of x=1t\dfrac{1}{t} and further, we will apply the integration formulae given as below-
∫0xdxx2+a2\int\limits_0^x {\dfrac{{dx}}{{{x^2} + {a^2}}}} =1atan⁡−1xa\dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a}

Complete step by step solution: Given definite integral in the question,
I=∫012dx(1+x2)1−x2\int\limits_0^{\dfrac{1}{2}} {\dfrac{{dx}}{{(1 + {x^2})\sqrt {1 - {x^2}} }}}
Putting the value of x = 1t\dfrac{1}{t} and differentiating it, dx = −dtt2\dfrac{{ - dt}}{{{t^2}}}, we will get
⇒∫012−dtt2(1+1t2)1−1t2\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,
⇒∫012−dtt2(t2+1t2)t2−1t2\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 ∫012−tdt(1+t2)t2−1\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 t2−1=u2{t^2} - 1 = {u^2}
⇒t2=u2+1\Rightarrow {t^2} = {u^2} + 1
Differentiating it we will find, tdt=udutdt = udu
Now, substituting all these values, the integral will become

⇒∫012−udu(1+u2+1)u2\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 ∫012−duu2+2\int\limits_0^{\dfrac{1}{2}} {\dfrac{{ - du}}{{{u^2} + 2}}}
Now, we will use the integration formulae which can be stated as
∫0xdxx2+a2\int\limits_0^x {\dfrac{{dx}}{{{x^2} + {a^2}}}} =1atan⁡−1xa\dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a}

So, our integral will be in terms of u will become,
I = 12tan⁡−1(−u)2∣012\left. {\dfrac{1}{{\sqrt 2 }}{{\tan }^{ - 1}}\dfrac{{( - u)}}{{\sqrt 2 }}} \right|_0^{\dfrac{1}{2}}
⇒−12tan⁡−1t2−12∣012\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 t2=1x2{t^2} = \dfrac{1}{{{x^2}}}and we will get ,
⇒−12tan⁡−11−x2x2∣012\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 = 12tan⁡−123\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 1(x2+a2)(x2−a2)\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
1(x2+a2)(x2−a2)\dfrac{1}{{({x^2} + {a^2})(\sqrt {{x^2} - {a^2}} )}}