can someone solve
∫ square root of (9-4x^2dx)
How about
$\displaystyle \int\sqrt{9-4x^2}\, dx=3\int \sqrt{1-\left(\tfrac{2}{3}x\right)^2}\, dx=\ldots$
Substituting $\displaystyle z := \tfrac{2}{3}x$, thus $\displaystyle dz = \tfrac{2}{3}dx$ and $\displaystyle dx = \tfrac{3}{2}dz$. This gives you
$\displaystyle =3\int \sqrt{1-z^2}\cdot \tfrac{3}{2}\,dz=\tfrac{9}{2}\int \sqrt{1-z^2}\, dz$
and, maybe, you kow how to solve this last integral...
Why don't you just use Wolfram Mathematica Online Integrator to check your result? - If it is not quite correct, then at least it is close...
I get (as does Wolfram's Integrator): $\displaystyle \int \sqrt{9-4x^2}\, dx=\tfrac{9}{4}\arcsin\left(\tfrac{2x}{3}\right)+{ \color{red}\tfrac{x}{2}}\sqrt{9-4x^2}+C$
Well, you have to enter the integrand first: I could not get a link to the result because that particular link got mangled when I tried to insert it into my post.it didn't show up in wolfram