Animated Solution for Mathematics - Definite Integration: For any integer $n$ the integral ${\int }_0^{\pi }{e}^{{\cos }^{2}x}{\cos }^3(2n+1)xdx$ has the value

Imagine you are standing before a formidable mathematical structure: the integral

At first glance, it looks like a chaotic collision of an exponential function and a high-frequency trigonometric wave. It is easy to feel intimidated, but in the world of JEE Advanced, complexity is often a mask for hidden elegance.

We are not here to brute-force this; we are here to uncover the symmetry that makes this monster collapse into nothingness.

When you see an integral from $0$ to $\pi$, your mind should immediately race to the King's Property:

This is not just a formula; it is a reflection. It tells us that the area under a curve is invariant if we flip the function across the midpoint of the interval.

Let us apply this to our integral by replacing $x$ with $(\pi -x)$. The new integral becomes:

Now, let us dissect the two components of our integrand. First, the exponential term: $e^{{\cos }^2(\pi -x)}$.

We know that $\cos (\pi -x)=-\cos x$. When we square this, the negative sign is swallowed by the even power: $(-\cos x{)}^2={\cos }^{2}x$. The exponential term is essentially a shield, remaining perfectly unchanged.

Next, consider the trigonometric term: ${\cos }^3((2n+1)(\pi -x))$. Expanding the angle gives us $(2n+1)\pi -(2n+1)x$.

Since $(2n+1)$ is always an odd integer, $(2n+1)\pi$ is an odd multiple of $\pi$. Trigonometry tells us that $\cos (\text{odd}\cdot \pi -\theta )=-\cos \theta$.

Because our term is cubed, the negative sign survives:

We have arrived at the moment of truth. Substituting these back into our integral, we get:

By pulling the negative sign out, we see that:

The integral on the right is exactly our original $I$. Thus, we have the simple, elegant equation $I=-I$.

Adding $I$ to both sides yields $2I=0$, which means $I=0$.

Geometrically, this means the function is anti-symmetric about $x=\frac{\pi }{2}$. The positive area from $0$ to $\frac{\pi }{2}$ is perfectly cancelled by the negative area from $\frac{\pi }{2}$ to $\pi$.