Animated Solution for Mathematics - Definite Integration: If ${\int }_0^{100\pi }\frac{{\sin }^{2}x}{e^{\left(\frac{x}{\pi }-\left[\frac{x}{\pi }\right]\right)}}dx=\frac{\alpha {\pi }^3}{1+4{\pi }^2},\alpha \in \mathbf{R}$ where $[x]$ is the greatest integer less than or equal to $x$, then the value of $\alpha$ is:

Welcome, fellow traveler on the journey of JEE Advanced mathematics. Today, we confront an integral that, at first glance, seems designed to intimidate:

It looks like a chaotic mess of exponentials, trigonometric functions, and greatest integer brackets. However, in mathematics, complexity is often just a mask for a hidden, elegant simplicity.

The first thing that should catch your eye is the exponent in the denominator: $\frac{x}{\pi }-[\frac{x}{\pi }]$. This is the textbook definition of the fractional part function, denoted as $\{\frac{x}{\pi }\}$.

By recognizing this, the denominator transforms from a terrifying expression into the much friendlier $e^{\{\frac{x}{\pi }\}}$. This is the first step in our victory: simplifying the integrand to:

Now, look at the limits of integration: $0$ to $100\pi$. We are not going to integrate this function one hundred times. Instead, we look for symmetry.

The numerator ${\sin }^{2}x$ has a period of $\pi$. The denominator, due to the properties of the fractional part function, also has a period of $\pi$. Because the entire function $f(x)$ is periodic with period $\pi$, we can use the property:

This allows us to reduce our massive integral to:

In the interval $[0,\pi ]$, the ratio $\frac{x}{\pi }$ lies between $0$ and $1$, meaning its greatest integer part is $0$. Thus, $\{\frac{x}{\pi }\}=\frac{x}{\pi }$. Our integral becomes:

To handle the ${\sin }^{2}x$, we use the half-angle identity ${\sin }^{2}x=\frac{1-\cos 2x}{2}$. This splits our problem into two manageable integrals:

The first is a simple exponential integral, and the second is a classic form solved using the standard formula:

After evaluating these integrals and carefully applying the limits, the terms collapse with satisfying precision. The sine terms vanish at the boundaries, and we are left with a clean algebraic expression.

When we combine everything, we arrive at the final result:

Comparing this to the required form, we identify the constants. Keep this in mind for your exam: whenever you see large limits and fractional parts, look for the period. It is almost always the key to the door.