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Question: Evaluate \[\int {x\sqrt {\dfrac{{{a^2} - {x^2}}}{{{a^2} + {x^2}}}} dx} \] A) \(\dfrac{1}{2}{a^2}{...

Evaluate ∫xa2−x2a2+x2dx\int {x\sqrt {\dfrac{{{a^2} - {x^2}}}{{{a^2} + {x^2}}}} dx}
A) 12a2cos⁡−1(x2a2)+12a4+x4+C\dfrac{1}{2}{a^2}{\cos ^{ - 1}}\left( {\dfrac{{{x^2}}}{{{a^2}}}} \right) + \dfrac{1}{2}\sqrt {{a^4} + {x^4}} + C
B) 12sin⁡−1(x2a2)+a4+x4+C\dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{{x^2}}}{{{a^2}}}} \right) + \sqrt {{a^4} + {x^4}} + C
C) 12a2sin⁡−1(x2a2)+12a4−x4+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) 12cos⁡−1(x2a2)+12a4−x4+C\dfrac{1}{2}{\cos ^{ - 1}}\left( {\dfrac{{{x^2}}}{{{a^2}}}} \right) + \dfrac{1}{2}\sqrt {{a^4} - {x^4}} + C

Explanation

Solution

Firstly, let x2=t{x^2} = t .
⇒xdx=dt2\Rightarrow xdx = \dfrac{{dt}}{2}
So, I=∫a2−ta2+tdt2I = \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=∫xa2−x2a2+x2dxI = \int {x\sqrt {\dfrac{{{a^2} - {x^2}}}{{{a^2} + {x^2}}}} dx}
Now, let x2=t{x^2} = t
⇒2xdx=dt ⇒xdx=dt2  \Rightarrow 2xdx = dt \\\ \Rightarrow xdx = \dfrac{{dt}}{2} \\\
So,

I=∫a2−ta2+tdt2 ⇒I=12∫a2−ta2+t×a2−ta2−tdt ⇒I=12∫(a2−t)2a4−t2dt ⇒I=12∫a2−ta4−t2dt ⇒I=12∫a2a4−t2dt−12∫ta4−t2dt ⇒I=a22sin⁡−1(ta2)−I1  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, I1=12∫ta4−t2dt{I_1} = \dfrac{1}{2}\int {\dfrac{t}{{\sqrt {{a^4} - {t^2}} }}dt}
Let, a4−t2=u{a^4} - {t^2} = u
⇒−2tdt=du ⇒tdt=−du2  \Rightarrow - 2tdt = du \\\ \Rightarrow tdt = - \dfrac{{du}}{2} \\\
⇒I1=−14∫duu\Rightarrow {I_1} = - \dfrac{1}{4}\int {\dfrac{{du}}{{\sqrt u }}}
=−14(u1212) =−u122  = - \dfrac{1}{4}\left( {\dfrac{{{u^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}}} \right) \\\ = - \dfrac{{{u^{\dfrac{1}{2}}}}}{2} \\\
Putting back u=a4−t2u = {a^4} - {t^2}
⇒I1=−12a4−t2\Rightarrow {I_1} = - \dfrac{1}{2}\sqrt {{a^4} - {t^2}}
Putting values of I1=−12a4−t2{I_1} = - \dfrac{1}{2}\sqrt {{a^4} - {t^2}} in I=a22sin⁡−1(ta2)−I1I = \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\dfrac{t}{{{a^2}}}} \right) - {I_1} , we get
⇒I=a22sin⁡−1(ta2)−(−12a4−t2)\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=x2t = {x^2}
⇒I=a22sin⁡−1(x2a2)+12a4−x4\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:
∫dxa2−x2=sin⁡−1xa+C ∫dxa2+x2=1atan⁡−1xa+C ∫dxx2+a2=ln⁡(x+x2+a2)+C ∫dxx2−a2=ln⁡(x+x2−a2)+C ∫dxa2−x2=12aln⁡a+xa−x+C ∫dxx2−a2=12aln⁡x−ax+a+C ∫a2−x2dx=x2a2−x2+a22sin⁡−1xa+C  \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 \\\