Question

$\int \frac{x^2}{(9-x^2{)}^{3/2}}dx=$

• A. $\frac{x}{\sqrt{9-x^2}}-si{n}^{-1}\frac{x}{3}+c$
• B. $\frac{x}{\sqrt{9-x^2}}-si{n}^{-1}\frac{x}{3}+c$
• C. $si{n}^{-1}\frac{x}{3}-\frac{x}{\sqrt{9-x^2}}+c$
• D. None of these

Answer 1

The correct option is A $\frac{x}{\sqrt{9-x^2}}-si{n}^{-1}\frac{x}{3}+c$

Put $x=3sin\theta \Rightarrow dx=3cos\theta d\theta$, therefore
$\int \frac{x^2}{(9-x^2{)}^{3/2}}dx=\int \frac{\left(9si{n}^2\theta \right)\left(3cos\theta \right)}{(9-9si{n}^2\theta {)}^{3/2}.}d\theta$
$=\int \frac{27si{n}^2\theta cos\theta }{27co{s}^3\theta }d\theta =\int ta{n}^2\theta d\theta =\int (se{c}^2\theta -1)d\theta$
$=tan\theta -\theta +c=tan\left\{si{n}^{-1}\left(\frac{x}{3}\right)\right\}-si{n}^{-1}\left(\frac{x}{3}\right)+c=tanta{n}^{-1}\left(\frac{\left(\frac{x}{3}\right)}{\sqrt{1-\left(\frac{x^2}{9}\right)}}\right)-si{n}^{-1}\left(\frac{x}{3}\right)+c=\frac{x}{\sqrt{9-x^2}}-si{n}^{-1}\left(\frac{x}{3}\right)+c$