Question

$\int \frac{4dx}{x^2\sqrt{4-9x^2}}$ is equal to

• 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)

$\frac{-\sqrt{4-9x^2}}{x}+C$

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