Mathematics

Evaluate $$\displaystyle \int_{}^{} {x\sqrt {\frac{{{a^2} - {x^2}}}{{{a^2} + {x^2}}}} dx  } $$


ANSWER

$$\displaystyle \frac{1}{2}{a^2}{\sin ^{ - 1}}\left( {\frac{{{x^2}}}{{{a^2}}}} \right) + \frac{1}{2}\sqrt {{a^4} - {x^4}} + C$$


SOLUTION
Consider $$i=\displaystyle \int{x\dfrac{\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}}}dx$$

 put $$x^2=t$$ $$\Rightarrow xdx=\dfrac{dt}{2}$$

$$I=\dfrac{1}{2}\displaystyle \int{\dfrac{\sqrt{a^2-t}}{\sqrt{a^2+t}}}dt=\dfrac{1}{2}\displaystyle \int{\dfrac{a^2-t}{\sqrt{a^4-t^2}}}dt$$

$$I=\dfrac{1}{2}\displaystyle \int{\dfrac{a^2dt}{\sqrt{a^4-t^2}}}-\dfrac{1}{2}\displaystyle \int{\dfrac{t}{\sqrt{a^4-t^2}}}dt$$

$$=\dfrac{a^2}{2}\sin^{-1}\dfrac{t}{a^2}+\dfrac{1}{4}\displaystyle \int{\dfrac{-2tdt}{\sqrt{a^4-t^2}}}$$ put $$a^4-t^2=u\\-2tdt=du$$

$$=\dfrac{a^2}{2}\sin^{-1}\dfrac{x^2}{a^2}+\boxed{\dfrac{1}{4}\displaystyle \int{\dfrac{du}{\sqrt u}}}\rightarrow \dfrac{1}{2}\sqrt u+c$$

$$=\dfrac{a^2}{2}\sin^{-1}\dfrac{x^2}{a^2}+\dfrac{1}{2}\sqrt{a^4-x^4}+c$$.
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Single Correct Medium Published on 17th 09, 2020
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