\(\int\sqrt{9-4x^2}\,dx\) is equal to (\(|x| \lt \frac{3}{2}\), C is a constant of integration)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{x}{2}\sqrt{9-4x^2}+\frac{9}{4}\sin^{-1}\left(\frac{2x}{3}\right)+C\)
Let \(u = 2x\), so \(du = 2\,dx\) and \(9-4x^2 = 9-u^2\):
\( \displaystyle \int\sqrt{9-4x^2}\,dx = \frac{1}{2}\int\sqrt{3^2-u^2}\,du \)
Using the standard integral \(\int\sqrt{a^2-u^2}\,du = \frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\sin^{-1}\frac{u}{a}+C\) with a = 3:
\( \displaystyle = \frac{1}{2}\left[\frac{u}{2}\sqrt{9-u^2}+\frac{9}{2}\sin^{-1}\frac{u}{3}\right]+C \)
Putting back u = 2x:
\( \displaystyle = \frac{1}{2}\left[x\sqrt{9-4x^2}+\frac{9}{2}\sin^{-1}\frac{2x}{3}\right]+C \)
\( \displaystyle = \frac{x}{2}\sqrt{9-4x^2}+\frac{9}{4}\sin^{-1}\left(\frac{2x}{3}\right)+C \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{x}{2}\sqrt{9-4x^2}+\frac{9}{8}\sin^{-1}\left(\frac{2x}{3}\right)+C\) — Incorrect. The \(\sin^{-1}\) coefficient is \(\frac{1}{2}\times\frac{9}{2} = \frac{9}{4}\), not \(\frac{9}{8}\).
Option (3) \(\displaystyle x\sqrt{9-4x^2}+\frac{9}{2}\sin^{-1}\left(\frac{2x}{3}\right)+C\) — Incorrect. It leaves out the factor \(\frac{1}{2}\) from \(dx = \frac{du}{2}\); it is twice the answer.
Option (4) \(\displaystyle\frac{x}{2}\sqrt{9-4x^2}+\frac{9}{4}\sin^{-1}\left(\frac{x}{3}\right)+C\) — Incorrect. The angle is \(\frac{u}{3} = \frac{2x}{3}\), not \(\frac{x}{3}\).