Animated Solution for Mathematics - Definite Integration: The area, enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ and the lines $x=0,x=\frac{\pi }{2}$ is:

Welcome, student. Today, we are not just solving a calculus problem; we are choreographing a dance between two mathematical functions. We are tasked with finding the area enclosed by $y_1=\sin x+\cos x$ and $y_2=|\cos x-\sin x|$ within the interval $[0,\frac{\pi }{2}]$.

This problem is a classic JEE Advanced challenge because it tests your ability to handle the modulus function—a function that demands respect and careful handling.

Imagine standing at the origin of a coordinate plane. To your right, stretching out to $\frac{\pi }{2}$, lies a region trapped between two distinct paths. The first path, $y_1=\sin x+\cos x$, is a smooth, continuous wave.

The second path, $y_2=|\cos x-\sin x|$, is more complex. Because of the absolute value bars, this curve is not just a simple wave; it is a 'V' shape or a cusp-like structure that reflects the behavior of the difference between sine and cosine.

Our goal is to calculate the area $A$ trapped between these two. The fundamental principle here is simple: the area between two curves is the integral of the upper curve minus the lower curve. Mathematically, we write this as:

The modulus function $|\cos x-\sin x|$ is a shape-shifter. It changes its definition based on whether the expression inside is positive or negative. We know that $\sin x$ and $\cos x$ are equal at $x=\frac{\pi }{4}$.

For the interval $x\in [0,\frac{\pi }{4}]$, $\cos x\ge \sin x$. Therefore, the modulus opens positively: $|\cos x-\sin x|=\cos x-\sin x$.

For the interval $x\in [\frac{\pi }{4},\frac{\pi }{2}]$, $\sin x$ becomes the dominant force, meaning $\cos x<\sin x$. In this region, the modulus must open with a negative sign to keep the result positive: $|\cos x-\sin x|=\sin x-\cos x$.

This is the 'surgery' phase of our calculus journey. We cannot integrate across the whole interval at once because the function definition changes at $\frac{\pi }{4}$. We must split our integral into two distinct parts.

Let us set up our two integrals. The total area $A$ is the sum of the area from $0$ to $\frac{\pi }{4}$ and the area from $\frac{\pi }{4}$ to $\frac{\pi }{2}$.

For the first part, $x\in [0,\frac{\pi }{4}]$:

Look closely at the integrand. When we distribute the negative sign, the $\cos x$ terms cancel out perfectly! We are left with $\sin x+\sin x$, which is simply $2\sin x$. The integral becomes:

Now, for the second part, $x\in [\frac{\pi }{4},\frac{\pi }{2}]$:

Again, watch the magic happen. The $\sin x$ terms cancel out, leaving us with $\cos x+\cos x$, which is $2\cos x$. The integral becomes:

We have reduced a seemingly terrifying problem into two simple, elegant integrals. Let us evaluate them.

For $A_1$, the integral of $2\sin x$ is $-2\cos x$. Evaluating from $0$ to $\frac{\pi }{4}$:

For $A_2$, the integral of $2\cos x$ is $2\sin x$. Evaluating from $\frac{\pi }{4}$ to $\frac{\pi }{2}$:

Notice the symmetry! Both parts yield the same value. Adding them together:

To match our options, we factor out $2\sqrt{2}$:

There you have it. We navigated the modulus, split the integral, simplified the terms, and arrived at the solution. Mathematics is not about memorizing formulas; it is about recognizing patterns and having the patience to peel back the layers of a problem.