Animated Solution for Mathematics - Definite Integration: Prove that for any positive integer $k$, $\frac{\sin 2kx}{\sin x}=2[\cos x+\cos 3x+\cdots +\cos (2k-1)x]$. Hence prove that ${\int }_0^{\pi /2}\sin 2kx\cot xdx=\frac{\pi }{2}$

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a beautiful trigonometric structure. In the JEE Advanced arena, you will often encounter expressions that look intimidating at first glance—like $\frac{\sin 2kx}{\sin x}$—but hide a simple, elegant truth.

Let us embark on this journey together.

The Identity

Imagine you are staring at the right-hand side of our identity: $2[\cos x+\cos 3x+\cdots +\cos (2k-1)x]$. It looks like a long, tedious sum.

But what if we could collapse it? The secret weapon here is the denominator on the left-hand side: $\sin x$. If we multiply our sum by $\frac{\sin x}{\sin x}$, we introduce the term $2\sin x\cos (nx)$ into every single part of the sum.

Why is this powerful? Because of the product-to-sum identity:

Let us apply this to the first term: $2\sin x\cos x=\sin 2x$. Now the second term:

Do you see the pattern emerging? The positive $\sin 2x$ from the second term is about to meet the negative $\sin 2x$ from the first term. This is the telescoping effect.

The Telescoping Magic

When we write out the full numerator, we get:

It is like a row of dominoes falling. The $\sin 2x$ cancels, the $\sin 4x$ cancels, and so on.

Every intermediate term vanishes into thin air, leaving only the very last term: $\sin 2kx$. Thus, we have proven that the sum is exactly $\frac{\sin 2kx}{\sin x}$. We have tamed the beast.

The Integration

Now, we turn our attention to the integral:

At first, this looks like a nightmare. How do you integrate $\cot x$ multiplied by a high-frequency sine wave?

But wait—we just proved that $\sin 2kx\cot x=\sin 2kx\frac{\cos x}{\sin x}=\left(\frac{\sin 2kx}{\sin x}\right)\cos x$. And we know exactly what that fraction is!

Substituting our identity, the integral becomes:

Distributing the $\cos x$, we get terms like $2{\cos }^{2}x$ and $2\cos (nx)\cos x$. Using the identity $2\cos A\cos B=\cos (A+B)+\cos (A-B)$, we transform the integrand into a simple sum of cosines:

The Final Victory

Integrating this is straightforward. The integral of $1$ is $x$, and the integral of $\cos (nx)$ is $\frac{\sin (nx)}{n}$.

When we evaluate this from $0$ to $\pi /2$, something magical happens. Every single sine term, $\sin (2x),\sin (4x),\dots ,\sin (2kx)$, becomes $\sin (\pi ),\sin (2\pi ),\dots ,\sin (k\pi )$ at the upper limit.

And what is the sine of any integer multiple of $\pi$? It is zero!

So, all those complex cosine terms vanish, leaving us with only the integral of the constant term $1$. The result is simply:

This, my friend, is the essence of JEE Advanced physics and mathematics. It is not about brute force; it is about finding the hidden pattern, simplifying the complex, and watching the chaos resolve into order. Keep practicing, keep visualizing, and keep falling in love with the process.