The integral $\int \frac{dx}{\sqrt{9-4x^2}}$ is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{1}{2} \sin^{-1} \left(\frac{2x}{3}\right) + c$
$\int \frac{1}{\sqrt{9-4x^2}} dx = \int \frac{1}{\sqrt{4 \left[ \left(\frac{3}{2}\right)^2 - x^2 \right]}} dx$
Using formula $\int \frac{1}{\sqrt{a^2-x^2}} = \sin^{-1} \frac{x}{a} + c$
$= \frac{1}{2} \sin^{-1} \frac{x}{3/2} + C = \frac{1}{2} \sin^{-1} \frac{2x}{3} + C$