Animated Solution for Mathematics - Definite Integration: The value of ${\int }_{-\pi /6}^{\pi /6}\left(\frac{\pi +4x^{11}}{1-\sin (|x|+\pi /6)}\right)dx$ is equal to:

A Masterclass in Integration

Imagine you are standing before a daunting integral. It looks complex, perhaps even intimidating, with its powers of $x$ and trigonometric functions.

But in the world of JEE Advanced, complexity is often a mask for elegance. Today, we are going to peel back that mask.

Look closely at the integral:

The very first thing that should catch your eye is the limits of integration: from $-\pi /6$ to $\pi /6$.

Whenever you see a symmetric interval like $[-a,a]$, your brain should immediately trigger a specific thought: Even and Odd functions. This is the hidden geometric reality of the problem; we aren't just calculating an area, we are looking for a balance.

To harness this symmetry, we split the integrand into two distinct parts:

Let's focus on the second integral. The numerator is $4x^{11}$, an odd power, while the denominator contains $|x|$, which is invariant under the transformation $x\to -x$.

Thus, the entire function is odd. As we know, the integral of an odd function over a symmetric interval is exactly zero. The area on the negative side perfectly cancels the area on the positive side, leaving us with only the first part.

Now we are left with the even function:

Since $h(-x)=h(x)$, we can use the property of even functions to simplify our limits:

Now, the absolute value is gone because $x\ge 0$. This is much cleaner, but that $x+\pi /6$ inside the sine is still a bit clunky.

Let's use a simple substitution: $u=x+\pi /6$, which implies $du=dx$. Our limits change from $[0,\pi /6]$ to $[\pi /6,\pi /3]$. The integral becomes:

This is a favorite concept of JEE examiners. How do we integrate $\frac{1}{1-\sin u}$?

We multiply the numerator and denominator by the conjugate, $(1+\sin u)$. The denominator becomes $1-{\sin }^{2}u$, which is ${\cos }^{2}u$.

Splitting the fraction gives us:

We are now in familiar territory. The integral of ${\sec }^{2}u$ is $\tan u$, and the integral of $\sec u\tan u$ is $\sec u$.

We evaluate the expression:

Substituting the limits:

Simplifying the arithmetic:

The $\sqrt{3}$ terms cancel out beautifully, leaving us with $2\pi \times 2=4\pi$.

And there you have it. The complexity dissolves, leaving behind a clean, elegant result of $4\pi$. Always remember: when the limits are symmetric, look for the symmetry in the function. It is the shortest path to the answer.