Question

Evaluate \[\int {x\sqrt {\dfrac{{{a^2} - {x^2}}}{{{a^2} + {x^2}}}} dx} \]
A) $\dfrac{1}{2}{a^2}{\cos ^{ - 1}}\left( {\dfrac{{{x^2}}}{{{a^2}}}} \right) + \dfrac{1}{2}\sqrt {{a^4} + {x^4}} + C$
B) $\dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{{x^2}}}{{{a^2}}}} \right) + \sqrt {{a^4} + {x^4}} + C$
C) $\dfrac{1}{2}{a^2}{\sin ^{ - 1}}\left( {\dfrac{{{x^2}}}{{{a^2}}}} \right) + \dfrac{1}{2}\sqrt {{a^4} - {x^4}} + C$
D) $\dfrac{1}{2}{\cos ^{ - 1}}\left( {\dfrac{{{x^2}}}{{{a^2}}}} \right) + \dfrac{1}{2}\sqrt {{a^4} - {x^4}} + C$

Answer 1

Hint:
Firstly, let ${x^2} = t$ .
 $\Rightarrow xdx = \dfrac{{dt}}{2}$
So, \[I = \int {\sqrt {\dfrac{{{a^2} - t}}{{{a^2} + t}}} \dfrac{{dt}}{2}} \].
Then, factorize the above integral.
Thus, finally on solving after factorization, we can get the correct answer.

Complete step by step solution:
Let \[I = \int {x\sqrt {\dfrac{{{a^2} - {x^2}}}{{{a^2} + {x^2}}}} dx} \]
Now, let ${x^2} = t$
 $
  \Rightarrow 2xdx = dt \\
  \Rightarrow xdx = \dfrac{{dt}}{2} \\
 $
So,
 \[
  I = \int {\sqrt {\dfrac{{{a^2} - t}}{{{a^2} + t}}} \dfrac{{dt}}{2}} \\
  \Rightarrow I = \dfrac{1}{2}\int {\sqrt {\dfrac{{{a^2} - t}}{{{a^2} + t}} \times \dfrac{{{a^2} - t}}{{{a^2} - t}}} dt} \\
  \Rightarrow I = \dfrac{1}{2}\int {\sqrt {\dfrac{{{{\left( {{a^2} - t} \right)}^2}}}{{{a^4} - {t^2}}}} dt} \\
  \Rightarrow I = \dfrac{1}{2}\int {\dfrac{{{a^2} - t}}{{\sqrt {{a^4} - {t^2}} }}dt} \\
  \Rightarrow I = \dfrac{1}{2}\int {\dfrac{{{a^2}}}{{\sqrt {{a^4} - {t^2}} }}dt} - \dfrac{1}{2}\int {\dfrac{t}{{\sqrt {{a^4} - {t^2}} }}dt} \\
  \Rightarrow I = \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\dfrac{t}{{{a^2}}}} \right) - {I_1} \\
 \]
Now, ${I_1} = \dfrac{1}{2}\int {\dfrac{t}{{\sqrt {{a^4} - {t^2}} }}dt} $
Let, ${a^4} - {t^2} = u$
 $
  \Rightarrow - 2tdt = du \\
  \Rightarrow tdt = - \dfrac{{du}}{2} \\
 $
 $\Rightarrow {I_1} = - \dfrac{1}{4}\int {\dfrac{{du}}{{\sqrt u }}} $
         $
   = - \dfrac{1}{4}\left( {\dfrac{{{u^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}}} \right) \\
   = - \dfrac{{{u^{\dfrac{1}{2}}}}}{2} \\
 $
Putting back $u = {a^4} - {t^2}$
 $\Rightarrow {I_1} = - \dfrac{1}{2}\sqrt {{a^4} - {t^2}} $
Putting values of ${I_1} = - \dfrac{1}{2}\sqrt {{a^4} - {t^2}} $ in \[I = \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\dfrac{t}{{{a^2}}}} \right) - {I_1}\] , we get
 \[\Rightarrow I = \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\dfrac{t}{{{a^2}}}} \right) - \left( { - \dfrac{1}{2}\sqrt {{a^4} - {t^2}} } \right)\]
Now, putting \[t = {x^2}\]
 \[\Rightarrow I = \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\dfrac{{{x^2}}}{{{a^2}}}} \right) + \dfrac{1}{2}\sqrt {{a^4} - {x^4}} \]

So, option (C) is correct.

Note:
Some properties of indefinite integrals:
 $
  \int {\dfrac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} = {\sin ^{ - 1}}\dfrac{x}{a} + C \\
  \int {\dfrac{{dx}}{{{a^2} + {x^2}}}} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + C \\
  \int {\dfrac{{dx}}{{\sqrt {{x^2} + {a^2}} }}} = \ln \left( {x + \sqrt {{x^2} + {a^2}} } \right) + C \\
  \int {\dfrac{{dx}}{{\sqrt {{x^2} - {a^2}} }}} = \ln \left( {x + \sqrt {{x^2} - {a^2}} } \right) + C \\
  \int {\dfrac{{dx}}{{{a^2} - {x^2}}}} = \dfrac{1}{{2a}}\ln \dfrac{{a + x}}{{a - x}} + C \\
  \int {\dfrac{{dx}}{{{x^2} - {a^2}}}} = \dfrac{1}{{2a}}\ln \dfrac{{x - a}}{{x + a}} + C \\
  \int {\sqrt {{a^2} - {x^2}} dx} = \dfrac{x}{2}\sqrt {{a^2} - {x^2}} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a} + C \\
 $