Animated Solution for Mathematics - Definite Integration: The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then $A^4$ is equal to

Unveiling the Area Between Waves

Welcome, future engineers and scientists! Today, we are not just solving a math problem; we are witnessing a beautiful, rhythmic dance between two of the most fundamental functions in the universe: the sine wave and the cosine wave.

Imagine these two functions as partners on a graph, constantly crossing paths, creating pockets of space between them. Our goal is to measure the area of one of these pockets. It is a classic JEE Advanced problem, and it is a perfect example of how calculus allows us to quantify the geometry of nature.

Finding the Intersection

Before we can measure the area, we must know where the 'pocket' begins and ends. We are looking for the points where the blue sine wave and the red cosine wave meet. Mathematically, this is the moment where their values are identical: $\sin x=\cos x$.

If we divide both sides by $\cos x$ (assuming $\cos x eq 0$), we arrive at the elegant identity $\tan x=1$. This is our key!

We know that $\tan x=1$ at $x=\frac{\pi }{4}$, and it repeats every $\pi$ radians. Thus, our consecutive intersection points are $x_1=\frac{\pi }{4}$ and $x_2=\frac{5\pi }{4}$. We have successfully locked down the 'where' of our problem.

The Geometry of the Region

Now, we must ask: which function is on top? This is a crucial step that many students overlook, leading to negative area results. In the interval $[\frac{\pi }{4},\frac{5\pi }{4}]$, we need to determine which curve is the 'ceiling' and which is the 'floor'.

If you visualize the unit circle or the graphs, you will see that in this specific interval, the sine curve is riding higher than the cosine curve. We can verify this with a simple test point, say $x=\frac{\pi }{2}$.

At this point, $\sin (\frac{\pi }{2})=1$ and $\cos (\frac{\pi }{2})=0$. Since $1>0$, we confirm that $\sin x\ge \cos x$. Therefore, our integrand must be $(\sin x-\cos x)$.

The Calculus of the Area

With our boundaries and our functions identified, we set up the definite integral for the area $A$:

This is where the magic happens. We apply the fundamental theorem of calculus. The antiderivative of $\sin x$ is $-\cos x$, and the antiderivative of $\cos x$ is $\sin x$.

Putting it all together, our antiderivative is:

Now, we evaluate this at the limits. For the upper limit, $x=\frac{5\pi }{4}$, both $\sin$ and $\cos$ are $-\frac{1}{\sqrt{2}}$. Substituting these values gives us:

Next, for the lower limit, $x=\frac{\pi }{4}$, both $\sin$ and $\cos$ are $\frac{1}{\sqrt{2}}$. Substituting these gives:

Subtracting the lower limit value from the upper limit value, we get $A=\sqrt{2}-(-\sqrt{2})=2\sqrt{2}$.

The Final Flourish

The problem asks for $A^4$. We have found $A=2\sqrt{2}$. Let us calculate the final value with precision:

Using the laws of exponents, we distribute the power:

And there we have it! The area raised to the fourth power is 64. You have navigated the intersection, mastered the geometry, executed the calculus, and arrived at the solution.