Statement-1: The value of the integral ∫_π/⁶π/³ dx / (1+√tan x) is equal to π/6. Statement-2: ∫ₐᵇ f(

Q: Why does the King's Property work?

Think of it as a geometric reflection. When you replace $x$ with $a+b-x$, you are effectively flipping the graph of the function about the midpoint of the interval $[a, b]$. The area under the curve remains unchanged because the shape is just mirrored. It is a beautiful way to simplify complex integrands.

Q: How do I know when to use the King's Property?

Whenever you see a definite integral with limits that sum to a 'nice' value like $\pi/2$ or $\pi$, and the integrand involves trigonometric functions like $\sin$, $\cos$, $\tan$, or $\cot$, the King's Property is almost always the intended path. It is a classic JEE signature.

Q: I got $\pi/12$, but the statement says $\pi/6$. Did I do something wrong?

You did everything perfectly! The fact that you arrived at $\pi/12$ is the correct mathematical result. The statement claiming it is $\pi/6$ is simply incorrect. Trust your derivation and stay confident in your process.

Question

Statement-1: The value of the integral $\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{	an x}}$ is equal to $\pi/6$. Statement-2: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$.

Options:

(A) Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

(C) Statement-1 is true; Statement-2 is false.

(D) Statement-1 is false; Statement-2 is true.

Sigma Percentile · Accepted Answer

### The Symmetry of the King's Property

Welcome, future engineer. Today, we are going to explore one of the most elegant tools in the calculus toolkit: the **King's Property**. This is not just a formula to memorize; it is a profound statement about symmetry in integration.

When you face an integral that looks like a tangled knot, the King's Property is often the key that unravels it.

### The Setup: Reading the Clues

Look at the integral in Statement-1:

$$I = \int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{	an x}}$$

At first glance, it looks intimidating. However, observe the limits: $\pi/6$ and $\pi/3$.

When you add them, you get:

$$\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi + 2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$$

This is the 'signature' of the King's Property. Whenever you see limits that sum to $\pi/2$ in a trigonometric integral, you should immediately consider the property:

$$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$$

### The Transformation

Let us apply this property by replacing $x$ with $(\frac{\pi}{2} - x)$. Our integral becomes:

$$I = \int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{	an(\frac{\pi}{2}-x)}}$$

Recall the trigonometric identity $	an(\frac{\pi}{2} - x) = \cot x$. Our integral transforms into:

$$I = \int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\cot x}}$$

To align this with our original integral, we convert $\cot x$ to $\frac{1}{	an x}$:

$$I = \int_{\pi/6}^{\pi/3} \frac{dx}{1+\frac{1}{\sqrt{	an x}}}$$

Multiplying the numerator and denominator by $\sqrt{	an x}$, we obtain:

$$I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{	an x}}{\sqrt{	an x} + 1} dx$$

### The Magic Addition

Here is where the beauty lies. We have two expressions for $I$. Let us add them together:

$$2I = \int_{\pi/6}^{\pi/3} \frac{1}{1+\sqrt{	an x}} dx + \int_{\pi/6}^{\pi/3} \frac{\sqrt{	an x}}{1+\sqrt{	an x}} dx$$

Because the limits are identical, we combine the integrands:

$$2I = \int_{\pi/6}^{\pi/3} \left( \frac{1 + \sqrt{	an x}}{1 + \sqrt{	an x}} ight) dx$$

The numerator and denominator cancel out, leaving us with the integral of $1$:

$$2I = \int_{\pi/6}^{\pi/3} 1 dx$$

### The Final Verdict

Evaluating this is straightforward:

$$2I = [x]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$$

Therefore, **$I = \frac{\pi}{12}$**.

Statement-1 claimed the value was $\pi/6$, which we have proven to be **false**. Since Statement-2 correctly identifies the King's Property, we conclude that **Statement-1 is false and Statement-2 is true**.

Animated Solution for Mathematics - Definite Integration: Statement-1: The value of the integral $\int_{π /6}^{π /3} \frac{d x}{1 + t a n x}$ is equal to $π /6$ . Statement-2: $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ .

Select Answer:

Visualized Solution

Analyzing Statement-2

Geometric Meaning of King's Property

Setting up Statement-1

Checking the Sum of Limits

Applying King's Property

Trigonometric Simplification

Converting to Tangent

Simplifying the New Integral

Adding Equation 1 and Equation 2

Simplifying the Sum

Evaluating the Integral

Final Calculation

Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Symmetry of the King's Property

The Setup

The Transformation

The Magic Addition

The Final Verdict

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