Animated Solution for Mathematics - Definite Integration: Find the area of the region bounded by the x-axis and the curves defined by $y=\tan x,-\frac{\pi }{3}\le x\le \frac{\pi }{3};y=\cot x,\frac{\pi }{6}\le x\le \frac{3\pi }{2}$.

Welcome, fellow traveler on the road to JEE excellence. Today, we aren't just solving a calculus problem; we are choreographing a dance between two of the most fundamental functions in trigonometry: the tangent and the cotangent.

Imagine standing on the Cartesian plane, looking at the region trapped between these two curves and the $x$-axis. It looks simple, but as with all things in JEE Advanced, the beauty lies in the precision of our steps.

First, we must orient ourselves. We are given $y=\tan x$ and $y=\cot x$. Our goal is to find the area bounded by these curves and the $x$-axis.

Before we dive into integration, we need to know where these two paths cross. We set them equal: $\tan x=\cot x$. Since $\cot x=\frac{1}{\tan x}$, this becomes ${\tan }^{2}x=1$.

In our interval of interest, this leads us to the elegant intersection point $x=\frac{\pi }{4}$. At this point, both functions meet at the height of $y=1$. This point is the pivot of our entire calculation.

Now, look at the geometry. The region starts at $x=\frac{\pi }{6}$ and ends at $x=\frac{\pi }{3}$. But notice the 'ceiling' of our region.

From $\frac{\pi }{6}$ to $\frac{\pi }{4}$, the curve $y=\tan x$ is the one defining the upper boundary. However, as we cross the threshold of $\frac{\pi }{4}$, the curve $y=\tan x$ begins to climb steeply, and it is now the curve $y=\cot x$ that defines the upper boundary until we reach $x=\frac{\pi }{3}$.

This is the 'trap'—if you try to integrate just one function, you will miss the physical reality of the shape. We must split our integral into two parts:

I know that seeing integrals can be daunting, but let's take a breath and look at the tools in our kit. We know that $\int \tan xdx=\ln |\sec x|$ and $\int \cot xdx=\ln |\sin x|$.

Let's evaluate $I_1$ first:

Now, for $I_2$, we apply the same logic:

We are almost there. The total area $A$ is the sum of these two values. When we add them together, we get a string of logarithms.

Using the property $\ln a+\ln b-\ln c-\ln d=\ln \left(\frac{a\cdot b}{c\cdot d}\right)$, we can collapse this entire expression into a single, beautiful term:

After simplifying the fractions inside the logarithm, we find that the expression reduces to $\ln (1.5)$. Calculating this gives us approximately $0.405$.

Take a moment to appreciate what you have done. You didn't just crunch numbers; you analyzed the behavior of functions, identified a geometric transition, and used the power of logarithmic properties to simplify a complex physical area into a single, clean value. This is the essence of physics and mathematics—finding order in the complexity.