Answer
$$\int^{3/4}_{-3/4}\frac{6dx}{\sqrt{9-4x^2}}dx=\pi$$
Work Step by Step
$$A=\int^{3/4}_{-3/4}\frac{6dx}{\sqrt{9-4x^2}}dx=6\int^{3/4}_{-3/4}\frac{dx}{\sqrt{3^2-(2x)^2}}$$
Take $u=2x$, which means $$du=2dx$$ $$dx=\frac{1}{2}du$$
For $x=3/4$, we have $u=3/2$
For $x=-3/4$, we have $u=-3/2$
Therefore, $$A=3\int^{3/2}_{-3/2}\frac{du}{\sqrt{3^2-u^2}}$$
We have $$\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\Big(\frac{x}{a}\Big)+C$$
Therefore, $$A=3\sin^{-1}\Big(\frac{u}{3}\Big)\Big]^{3/2}_{-3/2}$$ $$A=3\Big(\sin^{-1}\frac{\frac{3}{2}}{3}-\sin^{-1}\frac{-\frac{3}{2}}{3}\Big)$$ $$A=3\Big(\sin^{-1}\frac{1}{2}-\sin^{-1}\Big(-\frac{1}{2}\Big)\Big)$$ $$A=3\Big(\frac{\pi}{6}-\Big(-\frac{\pi}{6}\Big)\Big)=3\times\frac{2\pi}{6}=\pi$$
