Question

$\int x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx$ is equal to

• A. $\frac{a^2}{2}{\sin }^{-1}\left(\frac{x^2}{a^2}\right)+\frac{1}{2}\sqrt{a^4-x^4}+C$
• B. $\frac{a^2}{2}{\tan }^{-1}\left(\frac{x^2}{a^2}\right)+\frac{1}{2}\sqrt{a^4-x^4}+C$
• C. $\frac{a^2}{2}{\sin }^{-1}\left(\frac{x^2}{a^2}\right)+\frac{3}{2}\sqrt{a^2-x^2}+C$
• D. $\frac{a^2}{2}{\sin }^{-1}\left(\frac{x^2}{a^2}\right)+\frac{1}{2}\sqrt{a^2-x^2}+C$

Answer 1

The correct option is A $\frac{a^2}{2}{\sin }^{-1}\left(\frac{x^2}{a^2}\right)+\frac{1}{2}\sqrt{a^4-x^4}+C$

$I=\int x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx$
Put $x^2=t$
$\Rightarrow 2xdx=dt$
$\therefore I=\frac{1}{2}\int \sqrt{\frac{a^2-t}{a^2+t}}dt$
$=\frac{1}{2}\int \frac{a^2-t}{\sqrt{a^4-t^2}}dt$
$=\frac{1}{2}\int \frac{a^2}{\sqrt{a^4-t^2}}dt-\frac{1}{2}\int \frac{t}{\sqrt{a^4-t^2}}dt$
Put $a^4-t^2=u$
$\Rightarrow -2tdt=du$
$=\frac{1}{2}\int \frac{a^2}{\sqrt{(a^2{)}^2-t^2}}dt+\frac{1}{4}\int \frac{du}{\sqrt{u}}$
$=\frac{1}{2}{a}^2.{\sin }^{-1}\left(\frac{t}{a^2}\right)+\frac{1}{2}\sqrt{a^4-t^2}+C$
$=\frac{a^2}{2}{\sin }^{-1}\left(\frac{x^2}{a^2}\right)+\frac{1}{2}\sqrt{a^4-x^4}+C$