Animated Solution for Mathematics - Definite Integration: Which of the following statements is correct for the function $g(\alpha )$ for $\alpha \in \mathbb{R}$ such that $g(\alpha )={\int }_{\pi /6}^{\pi /3}\frac{{\sin }^{\alpha }x}{{\cos }^{\alpha }x+{\sin }^{\alpha }x}dx$

Unmasking the Integral

Welcome, fellow traveler on the path to JEE mastery. Today, we encounter a problem that serves as a perfect lesson in the philosophy of competitive mathematics.

We are presented with the function:

At first glance, the presence of $\alpha$ in the exponent feels like a wall. It suggests that the value of the integral might change wildly as $\alpha$ varies. But in the world of JEE, complexity is often just a mask for a deeper, more elegant simplicity.

A Geometric Insight

Whenever you see a definite integral with trigonometric functions where the limits sum to $\frac{\pi }{2}$, your mind should immediately jump to the King's Property:

Why is this so powerful? Because it allows us to 'flip' the interval. In our case, the limits are $a=\frac{\pi }{6}$ and $b=\frac{\pi }{3}$. Their sum is $a+b=\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{2}$. This is the key that unlocks the door.

Let us apply the property. We replace every instance of $x$ in our integrand with $(\frac{\pi }{2}-x)$. The integral becomes:

Now, we invoke the magic of complementary angles. We know that $\sin (\frac{\pi }{2}-x)=\cos x$ and $\cos (\frac{\pi }{2}-x)=\sin x$. Substituting these, our integral transforms into:

This is our second equation.

Now, we perform the move that makes this problem collapse. We add our original integral (Equation 1) and our transformed integral (Equation 2).

On the left side, we get $g(\alpha )+g(\alpha )=2g(\alpha )$. On the right side, because the limits and the denominators are identical, we can combine the numerators:

Look closely at that integrand. The numerator and the denominator are identical! They cancel out perfectly, leaving us with the integral of $1$ over the interval $[\frac{\pi }{6},\frac{\pi }{3}]$.

We are left with:

The integral of $1$ is simply the length of the interval, which is $\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}$. Thus, $2g(\alpha )=\frac{\pi }{6}$, which simplifies to:

The $\alpha$ has vanished! The function is a constant. Because $g(\alpha )=\frac{\pi }{12}$ for all $\alpha$, it is a horizontal line on the graph.

Since $g(\alpha )=g(-\alpha )$, the function is even. This journey shows us that even when a problem looks like a mountain, a single, well-placed insight can turn it into a flat, beautiful plain. Keep practicing, keep questioning, and keep falling in love with the logic.