Animated Solution for Mathematics - Definite Integration: If ${\int }_{\pi /6}^{\pi /3}\sqrt{1-\sin 2x}dx=\alpha +\beta \sqrt{2}+\gamma \sqrt{3}$, where $\alpha ,\beta$ and $\gamma$ are rational numbers, then $3\alpha +4\beta -\gamma$ is equal to _____

Welcome, fellow explorer of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a standard calculus exercise, but is actually a beautiful test of your attention to detail and your ability to visualize functions.

We are tasked with evaluating the integral:

When you see a square root containing a trigonometric expression, your internal alarm should ring. We need to simplify the radicand.

Recall the fundamental identity $1={\sin }^{2}x+{\cos }^{2}x$ and the double-angle formula $\sin 2x=2\sin x\cos x$. When we substitute these into our expression, something magical happens:

This is a perfect square! We can rewrite the integrand as:

Here is where many students stumble. We are tempted to write $\sqrt{(\sin x-\cos x{)}^2}=\sin x-\cos x$.

But stop! Remember that for any real number $a$, $\sqrt{a^2}=|a|$. This absolute value is not just a formality; it is the physical boundary of our problem.

We are integrating from $\pi /6$ to $\pi /3$. Within this range, the functions $\sin x$ and $\cos x$ cross paths. Specifically, at $x=\pi /4$, $\sin x=\cos x$.

Because the sign of $(\sin x-\cos x)$ changes at $\pi /4$, we must split our integral into two distinct regions.

In the first region, $x\in [\pi /6,\pi /4]$, the cosine function is larger than the sine function. Thus, $|\sin x-\cos x|=\cos x-\sin x$.

In the second region, $x\in [\pi /4,\pi /3]$, the sine function takes the lead, so $|\sin x-\cos x|=\sin x-\cos x$.

Our integral now becomes:

Now, we integrate term by term. For the first part:

For the second part:

Summing these together, we get:

By comparing this to the form $\alpha +\beta \sqrt{2}+\gamma \sqrt{3}$, we identify $\alpha =-1$, $\beta =2$, and $\gamma =-1$.

Finally, calculating $3\alpha +4\beta -\gamma$ gives us:

See how the complexity dissolved once we respected the modulus? You have navigated the trap, performed the integration, and arrived at the truth. Keep this level of precision in your toolkit, and no problem will ever be too daunting.