Answer to an integral is wrong I have an integral question, to which I found the same answer at Mathway and Integral but it is labeled wrong by someone:
Question: find $$\int\frac{1}{\sqrt{9-\frac{9x^2}{4}}}dx.$$
Answer: $$\frac{2}{3}\arcsin\left(\frac{x}{2}\right) + C.$$
Is the answer really wrong?

Your answer is definitely right!
Let $x=2 \sin \theta$, then $d x=2 \cos \theta d \theta$
$$
\begin{aligned}
I: &=\int \frac{1}{\sqrt{9-\frac{9x^{2}}{4}}} d x \\
&=\int \frac{1}{\sqrt{9-9 \sin ^{2} \theta}} \cdot 2 \cos \theta d \theta \\
&=\frac{2}{3} \int 1 d \theta \\
&=\frac{2}{3} \theta+C \\
&=\frac{2}{3} \sin ^{-1}\left(\frac{x}{2}\right)+C
\end{aligned}
$$
