Animated Solution for Mathematics - Definite Integration: If $I_n={\int }_{-\pi }^{\pi }\frac{\sin nx}{(1+{\pi }^x)\sin x}dx,n=0,1,2,...,then$

My dear student, welcome to a problem that looks like a monster but behaves like a lamb. When you first encounter the integral

it is natural to feel a shiver of intimidation. The limits are symmetric, from $-\pi$ to $\pi$, and that $(1+{\pi }^x)$ term in the denominator seems designed to ruin your day.

But in the world of JEE Advanced, whenever you see symmetric limits, you are being given a massive hint. The King's Property is your best friend here.

A Mirror Image

The King's Property states that

In our case, $a=-\pi$ and $b=\pi$, so $a+b=0$. This means we can replace $x$ with $-x$ without changing the value of the integral.

Let's see what happens when we do that:

Since $\sin (-\theta )=-\sin \theta$, the negative signs in the numerator and denominator cancel out. We are left with:

If you multiply the numerator and denominator by ${\pi }^x$, the denominator becomes $(1+{\pi }^x)$, and the integral transforms into:

Now, here is where the magic happens. We have two expressions for $I_n$. Let's add them together:

Factor out the $\sin nx$ in the numerator:

The $(1+{\pi }^x)$ terms cancel out perfectly! We are left with:

Since the integrand $\frac{\sin nx}{\sin x}$ is an even function, we can simplify this to:

Now we need to find the relationship between $I_n$ and $I_{n+2}$. Let's look at the difference:

Using the trigonometric identity $\sin C-\sin D=2\cos (\frac{C+D}{2})\sin (\frac{C-D}{2})$, the numerator becomes $2\cos (n+1)x\sin x$. The $\sin x$ terms cancel, leaving us with:

Integrating this, we get:

Since $\sin (n+1)\pi =0$ and $\sin 0=0$, the entire expression evaluates to zero. Thus, $I_n=I_{n+2}$.

This is a profound result! It tells us that all even-indexed integrals are equal to $I_0=0$, and all odd-indexed integrals are equal to $I_1=\pi$. You have just conquered a complex-looking problem by simply observing the symmetry.