## University Calculus: Early Transcendentals (3rd Edition)

$$\int^{2}_{-2}\frac{3dt}{4+3t^2}dt=\frac{\pi\sqrt3}{3}$$
$$A=\int^{2}_{-2}\frac{3dt}{4+3t^2}dt=3\int^{2}_{-2}\frac{dx}{2^2+(t\sqrt3)^2}$$ Take $u=t\sqrt3$, which means $$du=\sqrt3dt$$ $$dt=\frac{1}{\sqrt3}du$$ For $t=2$, we have $u=2\sqrt3$ For $t=-2$, we have $u=-2\sqrt3$ Therefore, $$A=\sqrt3\int^{2\sqrt3}_{-2\sqrt3}\frac{du}{2^2+u^2}$$ We have $$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\Big(\frac{x}{a}\Big)+C$$ Therefore, $$A=\sqrt3\times\frac{1}{2}\tan^{-1}\Big(\frac{u}{2}\Big)\Big]^{2\sqrt3}_{-2\sqrt3}$$ $$A=\frac{\sqrt3}{2}\Big(\tan^{-1}\sqrt3-\tan^{-1}\Big(-\sqrt3\Big)\Big)$$ $$A=\frac{\sqrt3}{2}\Big(\frac{\pi}{3}-\Big(-\frac{\pi}{3}\Big)\Big)$$ $$A=\frac{\sqrt3}{2}\times\frac{2\pi}{3}=\frac{\pi\sqrt3}{3}$$