Key Concepts and Formulas

• Property 1 (King Property): $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. A common application is $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. This property is useful for eliminating or simplifying the variable of integration when it appears linearly.
• Property 2: $\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx$ if $f(2a-x) = f(x)$. This property helps in reducing the upper limit of integration by half if the function exhibits symmetry about $x=a$.

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Step-by-Step Solution

Step 1: Define the integral and apply the King Property. Let the given integral be $I$. $I = \int\limits_0^\pi {xf\left( {\sin x} \right)dx} \quad (*)$ We apply the King Property with $a = \pi$. Replace $x$ with $(\pi - x)$: $I = \int\limits_0^\pi {(\pi - x)f\left( {\sin (\pi - x)} \right)dx}$ Using the trigonometric identity $\sin(\pi - x) = \sin x$, we get: $I = \int\limits_0^\pi {(\pi - x)f\left( {\sin x} \right)dx} \quad (**)$

Reasoning: The original integral contains a term $xf(\sin x)$. Applying the King Property allows us to transform the integrand in a way that, when combined with the original integral, can eliminate the '$x$' term, simplifying the problem.

Step 2: Combine the original and transformed integrals. Add equation $(*)$ and equation $(**)$: $I + I = \int\limits_0^\pi {xf\left( {\sin x} \right)dx} + \int\limits_0^\pi {(\pi - x)f\left( {\sin x} \right)dx}$ $2I = \int\limits_0^\pi {\left[ {xf\left( {\sin x} \right) + (\pi - x)f\left( {\sin x} \right)} \right]dx}$ Factor out $f(\sin x)$: $2I = \int\limits_0^\pi {\left[ {x + \pi - x} \right]f\left( {\sin x} \right)dx}$ $2I = \int\limits_0^\pi {\pi f\left( {\sin x} \right)dx}$ Since $\pi$ is a constant, we can take it out of the integral: $2I = \pi \int\limits_0^\pi {f\left( {\sin x} \right)dx}$

Reasoning: By adding the two forms of the integral, the '$x$' terms cancel out, leaving a simpler integral involving only $f(\sin x)$ multiplied by a constant.

Step 3: Apply Property 2 to simplify the integral $\int\limits_0^\pi {f\left( {\sin x} \right)dx}$. We consider the integral $J = \int\limits_0^\pi {f\left( {\sin x} \right)dx}$. This is in the form $\int_0^{2a} g(x) dx$ where $g(x) = f(\sin x)$ and $2a = \pi$, so $a = \pi/2$. We need to check if $g(2a-x) = g(x)$, which means checking if $f(\sin(\pi - x)) = f(\sin x)$. Since $\sin(\pi - x) = \sin x$, we have $f(\sin(\pi - x)) = f(\sin x)$. Thus, the condition for Property 2 is satisfied.

Applying Property 2: $\int\limits_0^\pi {f\left( {\sin x} \right)dx} = 2\int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx}$

Reasoning: The problem statement requires the integral to be in terms of $\int_0^{\pi/2} f(\sin x) dx$. Property 2 is used to change the upper limit of integration from $\pi$ to $\pi/2$ by exploiting the symmetry of the integrand.

Step 4: Substitute the result from Step 3 back into the expression for $2I$ and solve for $A$. From Step 2, we have $2I = \pi \int\limits_0^\pi {f\left( {\sin x} \right)dx}$. Substituting the result from Step 3: $2I = \pi \left[ 2\int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx} \right]$ $2I = 2\pi \int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx}$ Divide both sides by 2: $I = \pi \int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx}$ We are given that $I = A\int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx}$. Comparing our derived expression for $I$ with the given expression, we find: $A = \pi$

Reasoning: By substituting the simplified form of the integral back, we can directly compare the obtained expression with the given form to identify the value of $A$.

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Common Mistakes & Tips

• Incorrectly Applying Properties: Always ensure the conditions for properties like $\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx$ are met. In this case, verifying $\sin(\pi - x) = \sin x$ is crucial.
• Algebraic Errors: Be meticulous when adding integrals and simplifying expressions. Small mistakes can lead to an incorrect value for $A$.
• Forgetting the 'x': The presence of the '$x$' multiplier in the original integral is the primary motivation for using the King Property. Its elimination is the key step.

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Summary

This problem is solved by strategically applying two fundamental properties of definite integrals. First, the King Property ($\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$) is used to transform the integral $\int_0^\pi xf(\sin x) dx$ into an expression that, when combined with the original, eliminates the '$x$' term and results in $2I = \pi \int_0^\pi f(\sin x) dx$. Second, the property $\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx$ (when $f(2a-x) = f(x)$) is applied to the integral $\int_0^\pi f(\sin x) dx$, leveraging the identity $\sin(\pi - x) = \sin x$, to change the upper limit to $\pi/2$. This leads to $I = \pi \int_0^{\pi/2} f(\sin x) dx$, thus identifying $A = \pi$.

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