Animated Solution for Mathematics - Definite Integration: Let $f(x)={\int }_0^{x}g(t){\log }_e(\frac{1-t}{1+t})dt$, where $g$ is a continuous odd function. If ${\int }_{-\pi /2}^{\pi /2}(f(x)+\frac{x^2\cos x}{1+e^x})dx=(\frac{\pi }{\alpha }{)}^2-\alpha$, then $\alpha$ is equal to

Unlocking the Integral

Welcome, fellow traveler, to the heart of one of the most elegant problems in the JEE Advanced curriculum. When you first look at this integral, it might seem like a chaotic mess of logarithms, exponentials, and trigonometric functions.

In the world of competitive mathematics, complexity is often just a mask for a deeper, simpler truth. Today, we are going to peel back that mask.

The Parity Investigation

We begin with the function $f(x)={\int }_0^{x}g(t)\ln \left(\frac{1-t}{1+t}\right)dt$. The problem gives us a crucial piece of information: $g(t)$ is a continuous odd function, meaning $g(-t)=-g(t)$.

Now, look at the logarithmic term, $h(t)=\ln \left(\frac{1-t}{1+t}\right)$. If we replace $t$ with $-t$, we get:

Both $g(t)$ and $h(t)$ are odd. When you multiply two odd functions, the result is an even function.

Let $k(t)=g(t)\cdot h(t)$. Since $k(t)$ is even, its integral from $0$ to $x$, which is $f(x)$, must be an odd function. This is our first major breakthrough.

The Vanishing Act

Now, look at the main integral we need to solve:

We can split this into two parts. The first part is ${\int }_{-\pi /2}^{\pi /2}f(x)dx$.

Since we established that $f(x)$ is an odd function and the limits are symmetric ($-\pi /2$ to $\pi /2$), this entire integral is exactly $0$. The positive area and negative area cancel out perfectly.

The first term vanishes, leaving us with only the second part:

The King's Property

This is where the magic happens. We have a denominator of $1+e^x$, which is the classic signal to use King's Property: ${\int }_a^b\varphi (x)dx={\int }_a^b\varphi (a+b-x)dx$.

Here, $a+b=0$. Replacing $x$ with $-x$, we get:

By rewriting $e^{-x}$ as $1/e^x$, we obtain:

Now, add the two forms of $J$:

Since $x^2\cos x$ is an even function, we simplify the limits:

The Final Integration

We are left with a standard integral: ${\int }_0^{\pi /2}{x}^2\cos xdx$. We use Integration by Parts (ILATE rule).

Let $u=x^2$ and $dv=\cos xdx$. After two rounds of integration by parts, we arrive at the anti-derivative:

Evaluating this from $0$ to $\pi /2$:

Comparing this to the given form ${\left(\frac{\pi }{\alpha }\right)}^2-\alpha$, we see that $\alpha =2$. We have conquered the beast!