Animated Solution for Mathematics - Definite Integration: For $x\in \mathbb{R}$, let ${\tan }^{-1}(x)\in (-\frac{\pi }{2},\frac{\pi }{2})$. Then the minimum value of the function $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)={\int }_0^{x{\tan }^{-1}x}\frac{e^{t-\cos t}}{1+t^{2023}}dt$ is

Welcome, future IITian. Today, we are going to dismantle a problem that, at first glance, looks like it was designed to ruin your day. We have this integral:

It looks terrifying, doesn't it? You see that $2023$ in the exponent, you see the transcendental function in the numerator, and your brain immediately screams, 'Integration by parts! Substitution! Taylor series!'

But stop. Breathe. In the world of JEE Advanced, the most complex-looking problems are often the ones that reward the most elegant, simple, and intuitive thinking. Let's embark on this journey together.

Taming the Upper Limit

Before we even touch the integrand, let's look at the boundaries. The upper limit is $g(x)=x{\tan }^{-1}x$. This is the key to the entire problem.

Let's analyze its behavior across the real number line. If $x>0$, then ${\tan }^{-1}x$ is positive. A positive number times a positive number is positive.

If $x<0$, then ${\tan }^{-1}x$ is negative. A negative number times a negative number is positive. And if $x=0$, the product is zero.

Therefore, for all $x\in \mathbb{R}$, we have $g(x)\ge 0$. This is a massive realization. Our integral is always running from $0$ to some non-negative value. We are never integrating 'backwards'.

The Nature of the Integrand

Now, let's look at the 'monster' inside:

We need to know if this function is positive or negative. Look at the numerator: $e^{t-\cos t}$. The exponential function $e^u$ is strictly positive for any real $u$.

So, the numerator is always positive. Now look at the denominator: $1+t^{2023}$. Since we established that our integral only spans $t\ge 0$ (because $g(x)\ge 0$), we only care about positive $t$.

For $t\ge 0$, $t^{2023}$ is non-negative, so $1+t^{2023}\ge 1$. A positive numerator divided by a positive denominator is always positive. Thus, $h(t)>0$ for all $t\ge 0$.

The Geometric Intuition

Here is where the magic happens. A definite integral ${\int }_a^{b}h(t)dt$ represents the area under the curve $h(t)$ from $a$ to $b$.

We have established that $h(t)$ is always positive. We have established that our interval of integration is $[0,g(x)]$, where $g(x)\ge 0$.

This means we are calculating the area under a strictly positive curve. To minimize this area, we need the smallest possible interval. The smallest possible interval is when the upper limit equals the lower limit, i.e., $g(x)=0$.

The Elegant Solution

When $g(x)=0$, the integral becomes:

Since the area is always non-negative and we have found a point where it is zero, $0$ must be the minimum value.

We didn't need to integrate a single term. We didn't need to use complex substitution. We simply used the properties of the function to see the truth hidden in plain sight. This is the essence of JEE Advanced mathematics—seeing past the noise to the fundamental truth. Keep practicing this mindset, and you will conquer the exam.