Question

$\int \frac{dx}{(3+4x^2)\sqrt{(4-3x^2)}}=$

• A. $\frac{1}{\sqrt{75}}{\tan }^{-1}\left(\frac{5x}{3\sqrt{4-3x}}\right)+c$
• B. $\frac{1}{\sqrt{20}}{\tan }^{-1}\left(\frac{5x}{3\sqrt{4-3x}}\right)+c$
• C. $\frac{1}{5\sqrt{3}}{\tan }^{-1}\left(\frac{5x}{\sqrt{12-9x^2}}\right)+c$
• D. $\frac{1}{5\sqrt{3}}{\tan }^{-1}\left(\frac{5x}{\sqrt{12-9x^2}}\right)+c$

Answer 1

The correct option is C $\frac{1}{5\sqrt{3}}{\tan }^{-1}\left(\frac{5x}{\sqrt{12-9x^2}}\right)+c$

$I=\int \frac{dx}{(3+4x^2)\sqrt{(4-3x^2)}}$
Put $x=\frac{2}{\sqrt{3}}\sin t\Rightarrow \sqrt{4-3x^2}=2\cos t$
$\Rightarrow dx=\frac{2}{\sqrt{3}}\cos tdt$
$\Rightarrow \int \frac{\frac{2}{\sqrt{3}}\cos tdt}{(3+\frac{16}{3}{\sin }^{2}t)\sqrt{4-4{\sin }^{2}t}}$
$\Rightarrow I=\int \frac{\sqrt{3}dt}{9+16{\sin }^{2}t}$
$\Rightarrow I=\int \frac{\sqrt{3}{\sec }^{2}tdt}{9{\sec }^{2}t+16{\tan }^{2}t}$
$\Rightarrow I=\int \frac{\sqrt{3}{\sec }^{2}tdt}{9+25{\tan }^{2}t}$
$\tan t=r\Rightarrow {\sec }^{2}tdt=dr$
$\Rightarrow I=\int \frac{\sqrt{3}dr}{9+25r^2}=\frac{\sqrt{3}}{9}\int \frac{dr}{1+{\left(\frac{5r}{3}\right)}^2}=\frac{\sqrt{3}}{9}{\tan }^{-1}\left(\frac{5r}{3}\right)\times \frac{3}{5}$
$x=\frac{2}{\sqrt{3}}\sin t\Rightarrow \tan t=\frac{\sqrt{3}x}{\sqrt{4-3x^2}}$
$I=\frac{1}{5\sqrt{3}}{\tan }^{-1}\left(\frac{5x}{\sqrt{12-9x^2}}\right)+c$