Question

$\int\frac{4dx}{x^{2}\sqrt{4-9x^{2}}}=$

• A. $\sqrt{4-9x^{2}}+C$
• B. $\frac{-2}{3}\sqrt{4-9x^{2}}+C$
• C. $\frac{-\sqrt{4-9x^{2}}}{x}+C$
• D. $\frac{2}{3}\frac{\sqrt{4-9x^{2}}}{x}+C$
• E. $-3\sqrt{4-9x^{2}}+C$

Answer 1

Answer: (C)

Let $I=\int \frac{4 d x}{x^{2} \sqrt{4-9 x^{2}}}=4 \int \frac{d x}{x^{2} \sqrt{(2)^{2}-(3 x)^{2}}}$
put $ 3 x=2 \sin \theta$
$\Rightarrow \, 3 d x=2 \cos \theta \cdot d \theta$
Then, $I=4 \times \frac{2}{3} \int \frac{\cos \theta d \theta}{\frac{4}{9} \sin ^{2} \theta \sqrt{4-4 \sin ^{2} \theta}} $
$=6 \int \frac{\cos \theta d \theta}{\sin ^{2} \theta \cdot 2 \cos \theta}=3 \int cosec^{2} \theta d \theta $
$=-3 \cot \theta+C=-3 \frac{\sqrt{1-\sin ^{2} \theta}}{\sin \theta}+C$
$=\frac{-3 \sqrt{1-\left(\frac{3 x}{2}\right)^{2}}}{\left(\frac{3 x}{2}\right)}+C$
$=-\frac{2}{x} \frac{\sqrt{4-9 x^{2}}}{2}+C$
$=-\frac{\sqrt{4-9 x^{2}}}{x}+C$