Animated Solution for Mathematics - Definite Integration: The value of ${\int }_{-\pi }^{\pi }\frac{2y(1+\sin y)}{1+{\cos }^{2}y}dy$ is :

Imagine you are standing on the edge of a vast, symmetric landscape, looking at the integral:

At first glance, this expression might seem intimidating, a tangled mess of trigonometric functions and variables. But in the world of JEE Advanced, complexity is often just a mask for elegance.

Our journey begins with a simple, strategic decision: splitting the integral. By distributing the $2y$ in the numerator, we transform one daunting task into two distinct, manageable ones:

The Power of Parity

Now, we look at the first term, $g(y)=\frac{2y}{1+{\cos }^{2}y}$. Whenever you encounter limits from $-\pi$ to $\pi$, your mathematical intuition should immediately scream: "Check for symmetry!"

If we replace $y$ with $-y$, the denominator remains unchanged because ${\cos }^2(-y)={\cos }^{2}y$. However, the numerator becomes $-2y$. Thus, $g(-y)=-g(y)$.

This is the definition of an odd function. Geometrically, the area on the left side of the $y$-axis is the exact negative of the area on the right. When integrated over a symmetric interval, they cancel out to zero. Just like that, the first half of our problem vanishes into thin air!

The King's Gambit

We are left with the second integral, $h(y)=\frac{2y\sin y}{1+{\cos }^{2}y}$. Testing for symmetry again, we find that:

This is an even function! For even functions, the area from $-\pi$ to $\pi$ is simply twice the area from $0$ to $\pi$. Our integral simplifies to:

Now, we face the final boss: the $y$ in the numerator. This is where we invoke the King's Property: ${\int }_0^{a}f(y)dy={\int }_0^{a}f(a-y)dy$. By replacing $y$ with $\pi -y$, we get:

Since $\sin (\pi -y)=\sin y$ and ${\cos }^2(\pi -y)={\cos }^{2}y$, the integral becomes:

The Final Descent

When we add our original integral $I$ to this new form, the magic happens. The $y\sin y$ terms cancel out, leaving us with:

Dividing by two, we get:

This is a classic substitution problem. Let $t=\cos y$, so $dt=-\sin ydy$. Changing the limits from $[0,\pi ]$ to $[1,-1]$, we obtain:

Flipping the limits to remove the negative sign, we have:

The integral of $\frac{1}{1+t^2}$ is ${\tan }^{-1}t$. Evaluating this from $-1$ to $1$, we get:

Finally, $I=2\pi \cdot \frac{\pi }{2}={\pi }^2$. We have arrived at the destination: ${\pi }^2$. It is a beautiful, clean result that rewards our patience and our mastery of symmetry.