1. Key Concepts and Formulas

• King Property of Definite Integrals: For an integral $\int_a^b f(x) dx$, the property states that $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. This is particularly useful when the integrand has a term involving $x$ in the numerator and the limits of integration are symmetric or have a convenient sum.
• Trigonometric Identities:
  • $\cos^2 x + \sin^2 x = 1$
  • $\sin(2x) = 2 \sin x \cos x$
  • $\cos(2x) = 2 \cos^2 x - 1 \implies \cos^2 x = \frac{1 + \cos(2x)}{2}$
  • $\sin^2 x = \frac{1 - \cos(2x)}{2}$
• Substitution Rule for Definite Integrals: If $u = g(x)$, then $du = g'(x) dx$. The limits of integration also change from $x=a$ to $u=g(a)$ and from $x=b$ to $u=g(b)$.
• Standard Integral: $\int \frac{1}{a^2 - u^2} du = \frac{1}{2a} \ln \left| \frac{a+u}{a-u} \right| + C$ or $\int \frac{1}{u^2 - a^2} du = \frac{1}{2a} \ln \left| \frac{u-a}{u+a} \right| + C$. Also, $\int \sec^2 x dx = \tan x + C$.

2. Step-by-Step Solution

Let the given integral be $I$. $I = \int_0^\pi \frac{8 x \, dx}{4 \cos^2 x + \sin^2 x}$

Step 1: Apply the King Property We use the King Property, $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Here, $a=0$ and $b=\pi$. So, $a+b-x = 0+\pi-x = \pi-x$. Applying this to our integral: $I = \int_0^\pi \frac{8 (\pi-x) \, dx}{4 \cos^2 (\pi-x) + \sin^2 (\pi-x)}$ We know that $\cos(\pi-x) = -\cos x$ and $\sin(\pi-x) = \sin x$. Therefore, $\cos^2(\pi-x) = (-\cos x)^2 = \cos^2 x$ and $\sin^2(\pi-x) = (\sin x)^2 = \sin^2 x$. Substituting these back into the integral: $I = \int_0^\pi \frac{8 (\pi-x) \, dx}{4 \cos^2 x + \sin^2 x}$

Step 2: Add the original and modified integrals Now we have two expressions for $I$: Equation (1): $I = \int_0^\pi \frac{8 x \, dx}{4 \cos^2 x + \sin^2 x}$ Equation (2): $I = \int_0^\pi \frac{8 (\pi-x) \, dx}{4 \cos^2 x + \sin^2 x}$ Adding Equation (1) and Equation (2): $2I = \int_0^\pi \frac{8 x}{4 \cos^2 x + \sin^2 x} \, dx + \int_0^\pi \frac{8 (\pi-x)}{4 \cos^2 x + \sin^2 x} \, dx$ Since the denominators are the same, we can combine the numerators: $2I = \int_0^\pi \frac{8x + 8(\pi-x)}{4 \cos^2 x + \sin^2 x} \, dx$ $2I = \int_0^\pi \frac{8x + 8\pi - 8x}{4 \cos^2 x + \sin^2 x} \, dx$ $2I = \int_0^\pi \frac{8\pi}{4 \cos^2 x + \sin^2 x} \, dx$

Step 3: Simplify the denominator and evaluate the integral We can rewrite the denominator using $\cos^2 x = 1 - \sin^2 x$: $4 \cos^2 x + \sin^2 x = 4(1-\sin^2 x) + \sin^2 x = 4 - 4\sin^2 x + \sin^2 x = 4 - 3\sin^2 x$ Alternatively, we can write the denominator in terms of $\tan x$ by dividing the numerator and denominator by $\cos^2 x$: $2I = \int_0^\pi \frac{8\pi}{4 \cos^2 x + \sin^2 x} \, dx = 8\pi \int_0^\pi \frac{1}{4 \cos^2 x + \sin^2 x} \, dx$ Divide numerator and denominator by $\cos^2 x$: $2I = 8\pi \int_0^\pi \frac{\sec^2 x}{4 + \tan^2 x} \, dx$ Now, let's use the substitution $u = \tan x$. Then $du = \sec^2 x \, dx$. We need to change the limits of integration: When $x=0$, $u = \tan(0) = 0$. When $x=\pi$, $u = \tan(\pi) = 0$.

This substitution leads to an integral from 0 to 0, which is 0. This indicates that we might have made a simplification too early or there's a subtlety with the interval $[0, \pi]$. Let's re-examine the denominator and the interval.

The denominator is $4 \cos^2 x + \sin^2 x$. We can rewrite this as $3 \cos^2 x + (\cos^2 x + \sin^2 x) = 3 \cos^2 x + 1$. Or, as $4(1-\sin^2 x) + \sin^2 x = 4 - 3\sin^2 x$.

Let's go back to: $2I = 8\pi \int_0^\pi \frac{1}{4 \cos^2 x + \sin^2 x} \, dx$ We can express the denominator in terms of $\tan x$: $4 \cos^2 x + \sin^2 x = \cos^2 x (4 + \tan^2 x)$ So the integral becomes: $2I = 8\pi \int_0^\pi \frac{\sec^2 x}{4 + \tan^2 x} \, dx$ Let $u = \tan x$, so $du = \sec^2 x \, dx$. The limits are $x=0 \implies u=0$ and $x=\pi \implies u=0$. This implies the integral evaluates to 0, which is incorrect. The issue is that $\tan x$ has a discontinuity at $x=\pi/2$ within the interval $[0, \pi]$.

Let's use the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$. And $\int_0^{2a} f(x) dx = 0$ if $f(2a-x) = -f(x)$. Our integral is from $0$ to $\pi$. Let's check the integrand for symmetry. Let $g(x) = \frac{8x}{4 \cos^2 x + \sin^2 x}$. $g(\pi-x) = \frac{8(\pi-x)}{4 \cos^2(\pi-x) + \sin^2(\pi-x)} = \frac{8(\pi-x)}{4 \cos^2 x + \sin^2 x}$. So $I = \int_0^\pi g(x) dx = \int_0^\pi g(\pi-x) dx$. $2I = \int_0^\pi (g(x) + g(\pi-x)) dx = \int_0^\pi \frac{8x + 8(\pi-x)}{4 \cos^2 x + \sin^2 x} dx = \int_0^\pi \frac{8\pi}{4 \cos^2 x + \sin^2 x} dx$.

Now consider the integral $J = \int_0^\pi \frac{1}{4 \cos^2 x + \sin^2 x} \, dx$. We can use the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$. Here, $2a = \pi$, so $a = \pi/2$. Let $f(x) = \frac{1}{4 \cos^2 x + \sin^2 x}$. $f(\pi-x) = \frac{1}{4 \cos^2(\pi-x) + \sin^2(\pi-x)} = \frac{1}{4 \cos^2 x + \sin^2 x} = f(x)$. So, $J = 2 \int_0^{\pi/2} \frac{1}{4 \cos^2 x + \sin^2 x} \, dx$.

Divide numerator and denominator by $\cos^2 x$: $J = 2 \int_0^{\pi/2} \frac{\sec^2 x}{4 + \tan^2 x} \, dx$ Let $u = \tan x$. Then $du = \sec^2 x \, dx$. When $x=0$, $u = \tan(0) = 0$. When $x=\pi/2$, $u = \tan(\pi/2) \to \infty$.

So the integral becomes: $J = 2 \int_0^\infty \frac{1}{4 + u^2} \, du$ This is of the form $\int \frac{1}{a^2 + u^2} du = \frac{1}{a} \arctan \left( \frac{u}{a} \right)$. Here $a^2 = 4$, so $a=2$. $J = 2 \left[ \frac{1}{2} \arctan \left( \frac{u}{2} \right) \right]_0^\infty$ $J = \left[ \arctan \left( \frac{u}{2} \right) \right]_0^\infty$ $J = \arctan(\infty) - \arctan(0)$ $J = \frac{\pi}{2} - 0 = \frac{\pi}{2}$

Now, recall that $2I = 8\pi J$. $2I = 8\pi \left( \frac{\pi}{2} \right)$ $2I = 4\pi^2$ $I = \frac{4\pi^2}{2}$ $I = 2\pi^2$

3. Common Mistakes & Tips

• Division by Zero: Be cautious when dividing by $\cos^2 x$ if the interval includes $\pi/2$, where $\cos x = 0$. The King Property and the subsequent splitting of the integral into $[0, \pi/2]$ and $[\pi/2, \pi]$ (or using the property for $2a$) helps manage this.
• Incorrect Limit Transformation: Always remember to change the limits of integration when using the substitution rule. For definite integrals, this is crucial.
• Algebraic Simplification Errors: Double-check algebraic manipulations, especially when dealing with trigonometric functions and their squares.

4. Summary

The integral was evaluated using the King Property of definite integrals, which allowed us to eliminate the $x$ term in the numerator. By adding the original integral to the modified integral, we obtained a simpler integral involving a constant in the numerator. The denominator was then manipulated to enable a substitution, and the integral was evaluated over the appropriate limits. The property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ for even functions was used to handle the interval $[0, \pi]$ by splitting it into $[0, \pi/2]$.

The final answer is $\boxed{2 \pi^2}$. This corresponds to option (A).