Animated Solution for Mathematics - Definite Integration: Let the area of the region enclosed by the curve $y=\min \{\sin x,\cos x\}$ and the $x$ axis between $x=-\pi$ to $x=\pi$ be $A$. Then $A^2$ is equal to

Imagine you are standing before a blank canvas, ready to sketch the behavior of a function that seems simple but hides a beautiful complexity. We are tasked with finding the area enclosed by $y=\min \{\sin x,\cos x\}$ and the $x$-axis between $x=-\pi$ and $x=\pi$.

This is not just a calculus problem; it is a story of two waves, sine and cosine, constantly competing to see who can stay lower on the graph.

First, let us draw our players. We have the blue sine wave and the green cosine wave. In the interval $[-\pi ,\pi ]$, they cross paths multiple times.

To find these critical junction points, we set $\sin x=\cos x$, which leads us to the elegant equation $\tan x=1$. Within our specified range, this occurs at $x=-\frac{3\pi }{4}$ and $x=\frac{\pi }{4}$.

These points are the 'handover' moments where the minimum function switches its allegiance from one curve to the other. By identifying these, we have effectively partitioned our journey into manageable segments.

Now, we must calculate the area for each segment. Because we are looking for the area enclosed with the $x$-axis, we must be vigilant about signs. If the curve is below the $x$-axis, the integral will be negative, so we must negate it to find the positive area.

In the first segment, $[-\pi ,-\frac{3\pi }{4}]$, the sine curve is the minimum and it lies below the $x$-axis. We calculate:

Moving to the second segment, $[-\frac{3\pi }{4},-\frac{\pi }{2}]$, the cosine curve takes over as the minimum. It is also below the $x$-axis, so we calculate:

In the third segment, $[-\frac{\pi }{2},\frac{\pi }{4}]$, the curve is above the $x$-axis, and we integrate the cosine function:

Finally, in the fourth segment, $[\frac{\pi }{4},\pi ]$, the sine curve returns as the minimum and stays above the $x$-axis. We calculate:

Now, we bring all our pieces together. The total area $A$ is the sum of these four segments:

Look closely at the terms involving $\frac{1}{\sqrt{2}}$. They are perfectly balanced, with two negative and two positive instances, causing them to vanish entirely! We are left with $1+1+1+1=4$.

We have reached the end, but do not let your guard down! The question asks for $A^2$, not just $A$. This is a classic JEE trap designed to catch those who rush to the finish line.

Since $A=4$, we calculate $A^2=4^2=16$.

And there it is—our final, rock-solid answer of 16. You have successfully navigated the curves, handled the absolute values, and avoided the final trap.