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

Mastering the Definite Integral

Welcome, fellow traveler on the path to JEE Advanced excellence. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of trigonometry and algebra.

We are looking at the integral $I={\int }_{-\pi }^{\pi }\frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx$. When you first see this, it is natural to feel a bit overwhelmed.

The numerator has a product, the denominator has a squared cosine, and the limits are symmetric. But here is the secret: in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

The Art of Splitting

Our first move is to simplify the landscape. We have a product in the numerator: $2x(1+\sin x)$.

Let us distribute that $2x$ to see what we are really dealing with. We can rewrite our integral as the sum of two distinct parts:

By splitting the integral, we have transformed one intimidating problem into two manageable ones. This is the first step of a master strategist: divide and conquer.

Now, we look at these two pieces with the eyes of a mathematician.

The Power of Parity

Whenever you see symmetric limits like $[-\pi ,\pi ]$, your brain should immediately scream: "Check for symmetry!" Let us analyze the first integral, $f(x)=\frac{2x}{1+{\cos }^{2}x}$.

If we replace $x$ with $-x$, we get:

Because $f(-x)=-f(x)$, this is an odd function. Geometrically, this means the area on the left side of the y-axis is the exact negative of the area on the right side.

When you integrate an odd function over a symmetric interval, they cancel out perfectly to zero. Just like that, the first half of our problem vanishes into thin air!

Now, let us look at the second integral, $g(x)=\frac{2x\sin x}{1+{\cos }^{2}x}$. Replacing $x$ with $-x$ here gives us:

This is an even function! For even functions, the area on the left is identical to the area on the right.

So, we can simplify the integral by integrating from $0$ to $\pi$ and doubling the result:

The Magic of King's Property

We are left with $4{\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$. That $x$ in the numerator is still a nuisance.

This is where we invoke the legendary King's Property: ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$. Let us apply this by replacing $x$ with $\pi -x$:

Using our trigonometric identities, we know $\sin (\pi -x)=\sin x$ and ${\cos }^2(\pi -x)={\cos }^{2}x$. The integral becomes:

Now, for the grand finale. We have two expressions for $I$. Let us add them together:

Look at the numerator: $x\sin x+\pi \sin x-x\sin x$. The $x\sin x$ terms cancel out! We are left with:

The Final Integration

We have reduced a complex problem to a simple substitution. Let $t=\cos x$, so $dt=-\sin xdx$.

When $x=0,t=1$. When $x=\pi ,t=-1$. Our integral becomes:

This is the standard integral for the inverse tangent function. Evaluating it, we get:

Since ${\tan }^{-1}(1)=\frac{\pi }{4}$ and ${\tan }^{-1}(-1)=-\frac{\pi }{4}$, the result is:

And there you have it! Through the power of symmetry, the elegance of King's Property, and the precision of substitution, we have arrived at ${\pi }^2$.

Remember, in JEE Advanced, it is not just about calculating; it is about seeing the hidden structure. Keep practicing, keep questioning, and keep falling in love with the process!