Key Concepts and Formulas

1. King's Rule (Property of Definite Integrals): $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. This property is particularly useful for integrals with symmetric limits or when substituting $a+b-x$ simplifies the integrand.
2. Symmetric Limits Property: $\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$ if $f(x)$ is an even function ($f(-x)=f(x)$), and $0$ if $f(x)$ is an odd function ($f(-x)=-f(x)$).
3. Trigonometric Identities: $\sin^2 x + \cos^2 x = 1$ and $\sin(\frac{\pi}{2} - x) = \cos x$.

Step-by-Step Solution

Let the given integral be $I$. $I = \int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {1 + {2^x}}}} dx \quad \text{(Equation 1)}$

Step 1: Apply King's Rule. We use King's Rule with $a = -\pi/2$ and $b = \pi/2$. Thus, $a+b-x = (-\pi/2) + (\pi/2) - x = -x$. Replacing $x$ with $-x$ in the integrand: $I = \int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}(-x)} \over {1 + {2^{ - x}}}}} dx$ Reasoning: Applying King's Rule transforms the term $2^x$ in the denominator into $2^{-x}$, which can be manipulated to combine with the original integral.

Step 2: Simplify the integrand. We know that $\sin(-x) = -\sin x$, so $\sin^2(-x) = (-\sin x)^2 = \sin^2 x$. The denominator becomes $1 + 2^{-x} = 1 + \frac{1}{2^x} = \frac{2^x + 1}{2^x}$. Substituting these into the integral: $I = \int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {\left( {{{1 + {2^x}} \over {{2^x}}}} \right)}}} dx$ $I = \int\limits_{ - \pi /2}^{\pi /2} {{{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} dx \quad \text{(Equation 2)}$ Reasoning: This step rewrites the transformed integral into a form that can be easily added to the original integral.

Step 3: Add Equation 1 and Equation 2. Adding the two expressions for $I$: $I + I = \int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {1 + {2^x}}}} dx + \int\limits_{ - \pi /2}^{\pi /2} {{{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} dx$ $2I = \int\limits_{ - \pi /2}^{\pi /2} {\left( {{{{{\sin }^2}x} \over {1 + {2^x}}} + {{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \right)} dx$ Since the limits of integration are the same, we can combine the integrands: $2I = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}x\left( {{{1 + {2^x}} \over {1 + {2^x}}}} \right)} dx$ $2I = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}x} dx$ Reasoning: Adding the original integral and the transformed integral is a standard technique when using King's Rule. This step reveals a significant simplification, as the denominator term cancels out.

Step 4: Simplify the integral using symmetric limits. The integral is $2I = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}x} dx$. Let $f(x) = \sin^2 x$. We check if $f(x)$ is even or odd: $f(-x) = \sin^2(-x) = (-\sin x)^2 = \sin^2 x = f(x)$. Since $f(x)$ is an even function, we can use the symmetric limits property: $\int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}x} dx = 2\int\limits_0^{\pi /2} {{{\sin }^2}x} dx$ Substituting this back: $2I = 2\int\limits_0^{\pi /2} {{{\sin }^2}x} dx$ Dividing both sides by 2: $I = \int\limits_0^{\pi /2} {{{\sin }^2}x} dx$ Reasoning: By identifying $\sin^2 x$ as an even function, we reduce the integration interval from $[-\pi/2, \pi/2]$ to $[0, \pi/2]$, simplifying the calculation.

Step 5: Evaluate the integral $\int\limits_0^{\pi /2} {{{\sin }^2}x} dx$. Let $J = \int\limits_0^{\pi /2} {{{\sin }^2}x} dx$. We can use the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ with $a = \pi/2$: $J = \int\limits_0^{\pi /2} {{{\sin }^2}\left( {{\pi \over 2} - x} \right)} dx$ Using the identity $\sin(\pi/2 - x) = \cos x$: $J = \int\limits_0^{\pi /2} {{{\cos }^2}x} dx$ Now, add the two expressions for $J$: $J + J = \int\limits_0^{\pi /2} {{{\sin }^2}x} dx + \int\limits_0^{\pi /2} {{{\cos }^2}x} dx$ $2J = \int\limits_0^{\pi /2} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} dx$ Using the identity $\sin^2 x + \cos^2 x = 1$: $2J = \int\limits_0^{\pi /2} {1} dx$ $2J = {\left[ x \right]_0^{\pi /2}}$ $2J = {\pi \over 2} - 0$ $2J = {\pi \over 2}$ $J = {\pi \over 4}$ Reasoning: This step uses a common technique for integrals of $\sin^2 x$ or $\cos^2 x$ over $[0, \pi/2]$. By transforming $\sin^2 x$ to $\cos^2 x$ and adding the two forms, the integrand simplifies to 1, making the integration straightforward.

Step 6: Final Result. From Step 4, we established that $I = \int\limits_0^{\pi /2} {{{\sin }^2}x} dx$, which we have evaluated as $J$. Therefore, $I = J = {\pi \over 4}$.

Common Mistakes & Tips

• Incorrect application of King's Rule: Ensure that $a+b-x$ is correctly calculated and substituted.
• Errors in simplifying trigonometric functions: Be precise with identities like $\sin(-x)$ and $\sin(\pi/2 - x)$.
• Forgetting to check for even/odd functions: Applying the symmetric limits property requires verifying the function's parity. Incorrectly assuming parity can lead to a wrong answer.
• Algebraic errors when adding integrals: Careful manipulation is needed when combining terms in the integrand.

Summary

The integral was solved by applying King's Rule to transform the integrand, which allowed us to add the original integral and the transformed integral. This resulted in a simplified integral of $\sin^2 x$ over symmetric limits. By recognizing $\sin^2 x$ as an even function, we reduced the integration range to $[0, \pi/2]$. Finally, we evaluated the integral of $\sin^2 x$ over $[0, \pi/2]$ by using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ and the identity $\sin^2 x + \cos^2 x = 1$, yielding the value $\pi/4$.

The final answer is \boxed{{\pi \over 4}}. This corresponds to option (A).