3√34_3√2/4^48√9-4x^2,dx --- i.e.

Question

$\dfrac{3\sqrt3}4\displaystyle\int_{3\sqrt2/4}^{}\dfrac{48}{\sqrt{9-4x^2}}\,dx$ --- i.e. $\displaystyle\int_{3\sqrt2/4}^{3\sqrt3/4}\frac{48}{\sqrt{9-4x^2}}\,dx$ is equal to:

Accepted Answer

Using $\displaystyle\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}+C$ with $a=3/2$:

$$\int_{3\sqrt2/4}^{3\sqrt3/4}\frac{48}{\sqrt{9-4x^2}}\,dx = 24\left[\sin^{-1}\frac{2x}3\right]_{3\sqrt2/4}^{3\sqrt3/4} = 24\left[\sin^{-1}\frac{\sqrt3}2-\sin^{-1}\frac1{\sqrt2}\right]$$

$$= 24\left(\frac\pi3-\frac\pi4\right) = 24\cdot\frac{\pi}{12} = 2\pi$$

Solutions · 1