Q111 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics
$\int \frac{x^2}{\left(\sqrt{4-x^2}\right)^3} d x=$
- A$\frac{x^2}{\sqrt{4-x^2}}-\sin ^{-1}\left(\frac{x}{2}\right)+C$
- Correct answerB$\frac{x}{\sqrt{4-x^2}}-\tan ^{-1}\left(\frac{x}{\sqrt{4-x^2}}\right)+C$
- C$\frac{x}{\sqrt{4-x^2}}+\sin ^{-1}\left(\frac{2}{\sqrt{4-x^2}}\right)+C$
- D$\sqrt{4-x^2}-\tan ^{-1}\left(\frac{x}{2}\right)+C$
Solution
Given that,
$\int \frac{x^2}{\left(\sqrt{4-x^2}\right)^3} d x$ ...(i)
Substituting, $x=2 \sin \theta$ and $d x=2 \cos \theta d \theta$ Putting in Eq. (i),
$=\int \frac{(2 \sin \theta)^2}{\left.(\sqrt{4-(2 \sin \theta})^2\right)^3} 2 \cos \theta d \theta$
$=\int \frac{4 \times 2 \times \sin ^2 \theta \cos \theta d \theta}{\left(\sqrt{4-4 \sin ^2 \theta}\right)^3}=\int \frac{8 \sin ^2 \theta \cos \theta d \theta}{(2 \cos \theta)^3}$
$=\int \frac{\sin ^2 \theta \cos \theta}{\cos ^3 \theta} d \theta=\int \tan ^2 \theta d \theta$
$=\int\left(\sec ^2 \theta-1\right) d \theta=\int \sec ^2 \theta d \theta-\int d \theta$
$=\tan \theta-\theta+C$
where, $C$ is integration constant.
$\because \quad x=2 \sin \theta \text { or } \sin \theta=\frac{x}{2}$
$\therefore \quad \tan \theta=\frac{x}{\sqrt{4-x^2}}$ or $\theta=\tan ^{-1}\left(\frac{x}{\sqrt{4-x^2}}\right)$
$=\frac{x}{\sqrt{4-x^2}}-\tan ^{-1}\left(\frac{x}{\sqrt{4-x^2}}\right)+C$
Hence, the correct answer is option: 'B'.