Answer
$$\int^{1/5}_{-1/5}\frac{6dx}{\sqrt{4-25x^2}}dx=\frac{2\pi}{5}$$
Work Step by Step
$$A=\int^{1/5}_{-1/5}\frac{6dx}{\sqrt{4-25x^2}}dx=6\int^{1/5}_{-1/5}\frac{dx}{\sqrt{2^2-(5x)^2}}$$
Take $u=5x$, which means $$du=5dx$$ $$dx=\frac{1}{5}du$$
For $x=1/5$, we have $u=1$
For $x=-1/5$, we have $u=-1$
Therefore, $$A=\frac{6}{5}\int^{1}_{-1}\frac{du}{\sqrt{2^2-u^2}}$$
We have $$\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\Big(\frac{x}{a}\Big)+C$$
Therefore, $$A=\frac{6}{5}\sin^{-1}\Big(\frac{u}{2}\Big)\Big]^{1}_{-1}$$ $$A=\frac{6}{5}\Big(\sin^{-1}\frac{1}{2}-\sin^{-1}\Big(-\frac{1}{2}\Big)\Big)$$ $$A=\frac{6}{5}\Big(\frac{\pi}{6}-\Big(-\frac{\pi}{6}\Big)\Big)=\frac{6}{5}\times\frac{2\pi}{6}=\frac{2\pi}{5}$$
