Animated Solution for Mathematics - Definite Integration: Integrate ${\int }_0^{\pi }\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}dx$.

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to look at an integral that, at first glance, appears to be a nightmare of transcendental functions:

I know what you are thinking—how on earth do we integrate an exponential function of a cosine? If you try to use standard substitution, you will quickly find yourself in a loop of frustration. But here is the secret: in mathematics, when a problem looks impossible, it is usually because we are looking at it from the wrong angle.

A Strategic Masterstroke

In the world of definite integrals, there is a legendary tool known as the 'King's Property.' It states that for any continuous function $f(x)$ on the interval $[0,a]$, the integral ${\int }_0^{a}f(x)dx$ is identical to ${\int }_0^{a}f(a-x)dx$.

Why is this so powerful? Because it allows us to transform the variable $x$ into something that interacts beautifully with the limits of integration. In our case, $a=\pi$. Let us apply this transformation by replacing $x$ with $(\pi -x)$.

When we substitute $x=\pi -x$, our integral becomes:

Now, pause for a moment and recall your trigonometry. We know that $\cos (\pi -x)=-\cos x$.

This is the moment of truth. By substituting this identity, our integral transforms into:

Look closely at this new expression. The denominator is identical to our original integral, just rearranged!

This is where the magic happens. We now have two expressions for the same value $I$. Let us add them together:

Because the denominators are the same, we can combine the integrands into a single fraction:

Take a deep breath and look at that fraction. The numerator and the denominator are exactly the same! They cancel out perfectly to leave us with the integral of $1$. The terrifying exponential functions have vanished, leaving behind the simplest possible integrand:

Now, the path is clear. The integral of $1$ with respect to $x$ is simply $x$. Evaluating this from $0$ to $\pi$, we get:

So, we have $2I=\pi$. Dividing by $2$, we arrive at our final answer:

Isn't it beautiful? We started with a complex, intimidating expression and, through the clever application of symmetry, reduced it to a simple constant. This is the essence of JEE Advanced mathematics: it is not about brute force; it is about finding the elegant path that nature has hidden in plain sight.