Simplifying integral $\int_4^3 \sqrt{(x - 3)(4 - x)} dx$ by an easy approach So I have this Integral $$\int_4^3 \sqrt{(x - 3)(4 - x)} dx$$
I know I can easily evaluate it using the by first converting it into this form $$\int \sqrt{a^2 + x^2} dx$$ and then using the direct formula for this.
But since this one's a definite integral and while evaluating it's getting very long and taking time to solve. Also the probability of committing a mistake is high. 
I was wondering if there's an easy approach to evaluate such integrals without doing this much maths. 
This question appeared in my exam for for just 2 marks and it took me a long time to solve.  I don't think this much calculation is justified for just 2 marks. 
Kindly help me with an easy approach.
 A: 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 $2\sin\theta\cos\theta=\sin(2\theta)$
A: 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. 
A: A similar approach but without substitutions is to complete the square inside the radical to get
$$ \sqrt{-\left(x-\frac{7}{2}\right)^2 + \frac{1}{4}}. $$
Let $y$ equal this expression, square both sides, add $\left(x - \dfrac{7}{2}\right)^2$ to both sides.  Graph it.  What is it?  How does the integral relate to the graph?  (Be careful when answering that last question!)
