Animated Solution for Mathematics - Definite Integration: ${\int }_{-\pi }^{\pi }\frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx$ is

Welcome, future engineer. Today, we are not just solving a math problem; we are dissecting a beautiful piece of calculus architecture. We are looking at the integral:

At first glance, this looks like a monster. The numerator is a product, the denominator is a trigonometric mess, and the limits are intimidating. But in the JEE Advanced arena, complexity is often a mask for elegance. Let us peel it back.

The Power of Parity

Whenever you see limits like $[-\pi ,\pi ]$, your brain should immediately fire a signal: Symmetry! We are dealing with a symmetric interval.

The first step is to split the integral into two distinct parts by expanding the numerator:

Let us call the first part $f_1(x)$ and the second $f_2(x)$.

Look at $f_1(x)=\frac{2x}{1+{\cos }^{2}x}$. If we replace $x$ with $-x$, we get:

This is an odd function. The area under an odd function over a symmetric interval is exactly zero. The positive area on the right cancels the negative area on the left. Just like that, the first half of our problem vanishes into thin air.

The King's Gambit

Now we focus on $f_2(x)=\frac{2x\sin x}{1+{\cos }^{2}x}$. Testing for parity again, we find:

This is an even function. For even functions, we can simplify the integral to:

Now, we deploy the most powerful weapon in our arsenal: King's Property. This property states that ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$.

We replace $x$ with $(\pi -x)$. Our integral becomes:

Since $\sin (\pi -x)=\sin x$ and ${\cos }^2(\pi -x)={\cos }^{2}x$, the denominator remains untouched. We now have two versions of $I$. Adding them together, the $x$ and $-x$ terms in the numerator cancel out, leaving us with:

The variable $x$ is gone!

The Final Substitution

We are left with $I=2\pi {\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx$. This is a standard integral.

We use the substitution $t=\cos x$, which gives $dt=-\sin xdx$. Remember to change your limits: when $x=0$, $t=1$; when $x=\pi$, $t=-1$.

The integral transforms into:

By using the negative sign to flip the limits back to $[-1,1]$, we get:

This is the classic integral of ${\tan }^{-1}(t)$. Evaluating this, we get:

And there it is. Through symmetry, the King's Property, and a clean substitution, we have arrived at the final answer: ${\pi }^2$. This is the beauty of physics and math—no matter how complex the start, there is always a path to elegance.