Key Concepts and Formulas

• Property of Definite Integrals (King's Rule): For a continuous function $g(x)$ on the interval $[0, a]$, $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$
• Trigonometric Identity: $\sin(\pi - x) = \sin x$.
• Integral of $\sin x$: $\int \sin x dx = -\cos x + C$.

Step-by-Step Solution

Step 1: Define the integral and apply the King's Rule. Let the given integral be $I$. $I = \int_{0}^{\pi} f(x) \sin x dx \quad \ldots(1)$ We apply the King's Rule with $a = \pi$. This property allows us to replace $x$ with $(\pi - x)$ in the integrand without changing the value of the integral. $I = \int_{0}^{\pi} f(\pi-x) \sin(\pi-x) dx$ This step is taken because the given functional equation $f(x) + f(\pi-x) = \pi^2$ directly involves the term $f(\pi-x)$, and applying the King's Rule will introduce this term into our integral, enabling us to use the given relation.

Step 2: Utilize trigonometric identities. We use the trigonometric identity $\sin(\pi - x) = \sin x$. Substituting this into the transformed integral from Step 1: $I = \int_{0}^{\pi} f(\pi-x) \sin x dx \quad \ldots(2)$ This simplification is crucial as it makes the $\sin x$ factor common in both the original integral (1) and the transformed integral (2), which will facilitate combining them.

Step 3: Combine the integrals by adding Equation (1) and Equation (2). Adding the two expressions for $I$: $I + I = \int_{0}^{\pi} f(x) \sin x dx + \int_{0}^{\pi} f(\pi-x) \sin x dx$ $2I = \int_{0}^{\pi} [f(x) \sin x + f(\pi-x) \sin x] dx$ We can factor out $\sin x$ from the terms inside the integral: $2I = \int_{0}^{\pi} [f(x) + f(\pi-x)] \sin x dx$ This step is strategic because the sum of the integrands, $f(x) + f(\pi-x)$, directly matches the given functional relation.

Step 4: Apply the given functional relation. We are given that $f(x) + f(\pi-x) = \pi^2$. Substituting this into the expression for $2I$: $2I = \int_{0}^{\pi} (\pi^2) \sin x dx$ Since $\pi^2$ is a constant, we can take it out of the integral: $2I = \pi^2 \int_{0}^{\pi} \sin x dx$ This step is the key to simplifying the problem, as the unknown function $f(x)$ has been eliminated, leaving a standard integral.

Step 5: Evaluate the simplified integral. We now evaluate the definite integral of $\sin x$ from $0$ to $\pi$: $\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi}$ $= (-\cos \pi) - (-\cos 0)$ $= (-(-1)) - (-1)$ $= 1 + 1 = 2$ This is a standard evaluation of a trigonometric definite integral. Remember that $\cos \pi = -1$ and $\cos 0 = 1$.

Step 6: Solve for $I$. Substitute the value of the integral back into the equation for $2I$: $2I = \pi^2 (2)$ $2I = 2\pi^2$ Dividing both sides by 2 to find the value of $I$: $I = \pi^2$

Common Mistakes & Tips

• Incorrect Application of King's Rule: Ensure you correctly substitute $(\pi-x)$ for every instance of $x$ in the integrand.
• Trigonometric Errors: Be careful with signs and values when evaluating trigonometric functions at the limits, especially $\cos \pi$ and $\cos 0$.
• Algebraic Slip-ups: Double-check the addition of integrals and the final division to solve for $I$.

Summary

The problem is solved by strategically applying the King's Rule for definite integrals, which states that $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$. By applying this rule to the integral $I = \int_{0}^{\pi} f(x) \sin x dx$, we obtained an equivalent expression $I = \int_{0}^{\pi} f(\pi-x) \sin(\pi-x) dx$. Using the identity $\sin(\pi-x) = \sin x$, we then added the original and transformed integrals. This allowed us to utilize the given functional relation $f(x) + f(\pi-x) = \pi^2$ to simplify the integrand to a constant $\pi^2$. The resulting integral $\pi^2 \int_{0}^{\pi} \sin x dx$ was evaluated to $2\pi^2$. Finally, solving for $I$ from $2I = 2\pi^2$ yielded $I = \pi^2$.

The final answer is $\boxed{\pi^{2}}$.