Simplifying integral $\int_4^3 \sqrt{(x - 3)(4 - x)} dx$ by an easy approach

Let $x=3.5+t$. We are finding $\int_{1/2}^{-1/2}\sqrt{(1/2)^2-t^2}\,dt$, which we recognize as (the negative of) a familiar area.

Here a elaboration.

Given integral is $\int_4^3 \sqrt{(x-3)(4-x)} \;\mathbb{d}x$

if $x=3\sin^2\theta+4\cos^2\theta$, then integral is from $0\leq\theta\leq\frac{\pi}{2}$.

$\mathbb{d}x=(3\times 2\times sin\theta\times\cos\theta-4\times2\times\cos\theta\times\sin\theta)\mathbb{d}\theta$

$\mathbb{d}x=-2\sin\theta\cos\theta\;\mathbb{d}\theta$

$$I=\int_0^{\frac{\pi}{2}}\cos\theta\times\sin\theta\times(-2\sin\theta\cos\theta)\mathbb{d}\theta$$

If you know beta function, it's a pretty straightforward integral but even if you don't, just use $2\sin\theta\cos\theta=\sin(2\theta)$