Key Concepts and Formulas

• Property of Definite Integrals: For a definite integral $\int_a^b f(x) \,dx$, the value remains the same if $x$ is replaced by $a+b-x$. Mathematically, $\int_a^b f(x) \,dx = \int_a^b f(a+b-x) \,dx$.
• Symmetric Limits Property: When the limits of integration are symmetric, i.e., from $-A$ to $A$, the property $\int_{-A}^A f(x) \,dx = \int_{-A}^A f(-x) \,dx$ is particularly useful.
• Integral of Odd/Even Functions:
  • If $f(x)$ is an odd function ($f(-x) = -f(x)$), then $\int_{-A}^A f(x) \,dx = 0$.
  • If $f(x)$ is an even function ($f(-x) = f(x)$), then $\int_{-A}^A f(x) \,dx = 2\int_0^A f(x) \,dx$.

Step-by-Step Solution

Let the given integral be denoted by $I$: $I = \int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$

Step 1: Apply the property of definite integrals with symmetric limits. The limits of integration are from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. We can use the property $\int_a^b f(x) \,dx = \int_a^b f(a+b-x) \,dx$. Here, $a = -\frac{\pi}{2}$ and $b = \frac{\pi}{2}$, so $a+b = 0$. Replacing $x$ with $a+b-x = 0-x = -x$ in the integrand: The integrand is $f(x) = \frac{\cos^2 x}{1+3^x}$. Then $f(-x) = \frac{\cos^2 (-x)}{1+3^{-x}}$. Since $\cos(-x) = \cos x$, we have $\cos^2(-x) = \cos^2 x$. So, $f(-x) = \frac{\cos^2 x}{1+3^{-x}}$. We can rewrite $1+3^{-x}$ as $1 + \frac{1}{3^x} = \frac{3^x+1}{3^x}$. Therefore, $f(-x) = \frac{\cos^2 x}{\frac{3^x+1}{3^x}} = \frac{3^x \cos^2 x}{1+3^x}$.

Using the property $\int_{-A}^A f(x) \,dx = \int_{-A}^A f(-x) \,dx$, we get: I = \int\limits_{ - \pi /2}^{\pi /2} {{{3^x}{{\cos }^2}x} \over {1 + {3^x}}}} dx Let's call this equation (2). The original integral is equation (1).

Step 2: Add the two expressions for the integral. We now have two expressions for $I$: (1) $I = \int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$ (2) I = \int\limits_{ - \pi /2}^{\pi /2} {{{3^x}{{\cos }^2}x} \over {1 + {3^x}}}} dx

Adding (1) and (2): 2I = \int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx + \int\limits_{ - \pi /2}^{\pi /2} {{{3^x}{{\cos }^2}x} \over {1 + {3^x}}}} dx Since the limits of integration are the same, we can combine the integrands: 2I = \int\limits_{ - \pi /2}^{\pi /2} {\left( {{{{{\cos }^2}x} \over {1 + {3^x}}}} + {{{3^x}{{\cos }^2}x} \over {1 + {3^x}}}} \right)} dx $2I = \int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x + 3^x{{\cos }^2}x} \over {1 + {3^x}}}} dx$ Factor out $\cos^2 x$ from the numerator: 2I = \int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x}(1 + {3^x})} \over {1 + {3^x}}}} dx

Step 3: Simplify the integrand and evaluate the integral. The term $(1+3^x)$ cancels out from the numerator and denominator: $2I = \int\limits_{ - \pi /2}^{\pi /2} {{{\cos }^2}x} dx$ Now we need to evaluate the integral of $\cos^2 x$. We use the double angle identity: $\cos(2x) = 2\cos^2 x - 1$, which means $\cos^2 x = \frac{1+\cos(2x)}{2}$. $2I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{1+\cos(2x)}{2}} dx$ $2I = \frac{1}{2} \int\limits_{ - \pi /2}^{\pi /2} {(1+\cos(2x))} dx$ Now, integrate term by term: $\int 1 \,dx = x$ $\int \cos(2x) \,dx = \frac{1}{2}\sin(2x)$ So, the integral becomes: $2I = \frac{1}{2} \left[ x + \frac{1}{2}\sin(2x) \right]_{ - \pi /2}^{\pi /2}$ Evaluate at the limits: $2I = \frac{1}{2} \left[ \left( \frac{\pi}{2} + \frac{1}{2}\sin\left(2 \cdot \frac{\pi}{2}\right) \right) - \left( -\frac{\pi}{2} + \frac{1}{2}\sin\left(2 \cdot -\frac{\pi}{2}\right) \right) \right]$ $2I = \frac{1}{2} \left[ \left( \frac{\pi}{2} + \frac{1}{2}\sin(\pi) \right) - \left( -\frac{\pi}{2} + \frac{1}{2}\sin(-\pi) \right) \right]$ Since $\sin(\pi) = 0$ and $\sin(-\pi) = 0$: $2I = \frac{1}{2} \left[ \left( \frac{\pi}{2} + 0 \right) - \left( -\frac{\pi}{2} + 0 \right) \right]$ $2I = \frac{1}{2} \left[ \frac{\pi}{2} - (-\frac{\pi}{2}) \right]$ $2I = \frac{1}{2} \left[ \frac{\pi}{2} + \frac{\pi}{2} \right]$ $2I = \frac{1}{2} \left[ \pi \right]$ $2I = \frac{\pi}{2}$

Step 4: Solve for I. $I = \frac{\pi}{4}$

Let's recheck the calculation carefully. The correct answer is stated as A, which is $2\pi$. My result is $\frac{\pi}{4}$. There must be an error in my steps or understanding.

Let's re-examine the question and options. The options are: (A) $2\pi$ (B) $\frac{\pi}{2}$ (C) $4\pi$ (D) $\frac{\pi}{4}$

My calculation leads to $\frac{\pi}{4}$, which is option (D). However, the provided correct answer is (A) $2\pi$. This indicates a significant discrepancy. Let me review the problem and the typical application of the property.

The property $\int_{-A}^A f(x) dx = \int_{-A}^A f(a+b-x) dx$ is correctly applied. $f(x) = \frac{\cos^2 x}{1+3^x}$ $f(-x) = \frac{\cos^2(-x)}{1+3^{-x}} = \frac{\cos^2 x}{1+1/3^x} = \frac{3^x \cos^2 x}{1+3^x}$. $I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1+3^x} dx$ $I = \int_{-\pi/2}^{\pi/2} \frac{3^x \cos^2 x}{1+3^x} dx$ $2I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x (1+3^x)}{1+3^x} dx = \int_{-\pi/2}^{\pi/2} \cos^2 x dx$. $\int_{-\pi/2}^{\pi/2} \cos^2 x dx = \int_{-\pi/2}^{\pi/2} \frac{1+\cos(2x)}{2} dx = \frac{1}{2} [x + \frac{1}{2}\sin(2x)]_{-\pi/2}^{\pi/2}$ $= \frac{1}{2} [(\frac{\pi}{2} + \frac{1}{2}\sin(\pi)) - (-\frac{\pi}{2} + \frac{1}{2}\sin(-\pi))]$ $= \frac{1}{2} [\frac{\pi}{2} - (-\frac{\pi}{2})] = \frac{1}{2} [\pi] = \frac{\pi}{2}$. So, $2I = \frac{\pi}{2}$, which means $I = \frac{\pi}{4}$.

There seems to be an issue with the provided correct answer, as my derivation consistently leads to $\frac{\pi}{4}$. Let me assume, for the sake of demonstrating the process to match a potential correct answer, that there might be a misunderstanding or a typo in the question or options. However, based on standard mathematical procedures, the result is $\frac{\pi}{4}$.

Let's consider if the question was different. For example, if the integral was $\int_{0}^{\pi/2} \cos^2 x dx$. $\int_{0}^{\pi/2} \cos^2 x dx = \int_{0}^{\pi/2} \frac{1+\cos(2x)}{2} dx = \frac{1}{2} [x + \frac{1}{2}\sin(2x)]_{0}^{\pi/2}$ $= \frac{1}{2} [(\frac{\pi}{2} + \frac{1}{2}\sin(\pi)) - (0 + \frac{1}{2}\sin(0))] = \frac{1}{2} [\frac{\pi}{2}] = \frac{\pi}{4}$.

The integral $\int_{-\pi/2}^{\pi/2} \cos^2 x dx$ is indeed $\frac{\pi}{2}$.

If $2I = \frac{\pi}{2}$, then $I = \frac{\pi}{4}$. If the correct answer is indeed $2\pi$, then $2I = 4\pi$. This would imply $\int_{-\pi/2}^{\pi/2} \cos^2 x dx = 4\pi$, which is incorrect.

Let's assume there's a mistake in the question or the given answer. If we strictly follow the problem statement and standard calculus, the answer is $\frac{\pi}{4}$.

However, if the question intended the answer to be $2\pi$, there must be a factor of 8 difference or some other error in the problem setting.

Let's re-read the question and options. Question: The value of $\int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$ is : Options: (A) $2\pi$ (B) $\frac{\pi}{2}$ (C) $4\pi$ (D) $\frac{\pi}{4}$

My calculation is correct and yields $\frac{\pi}{4}$. This matches option (D).

There is a possibility that the provided "Correct Answer: A" is erroneous. I will proceed with the derivation that leads to $\frac{\pi}{4}$.

Step-by-Step Solution (Revised based on my derivation)

Let the given integral be $I$: $I = \int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$

Step 1: Utilize the property $\int_a^b f(x) \,dx = \int_a^b f(a+b-x) \,dx$. Here $a = -\frac{\pi}{2}$ and $b = \frac{\pi}{2}$, so $a+b = 0$. Let $f(x) = \frac{\cos^2 x}{1+3^x}$. Then $f(-x) = \frac{\cos^2(-x)}{1+3^{-x}} = \frac{\cos^2 x}{1 + \frac{1}{3^x}} = \frac{\cos^2 x}{\frac{3^x+1}{3^x}} = \frac{3^x \cos^2 x}{1+3^x}$. Applying the property, we get a second expression for $I$: I = \int\limits_{ - \pi /2}^{\pi /2} {{{3^x}{{\cos }^2}x} \over {1 + {3^x}}}} dx

Step 2: Add the two expressions for $I$. 2I = \int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx + \int\limits_{ - \pi /2}^{\pi /2} {{{3^x}{{\cos }^2}x} \over {1 + {3^x}}}} dx 2I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\cos }^2}x} + {3^x}{{\cos }^2}x}{1 + {3^x}}}} dx 2I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\cos }^2}x}(1 + {3^x})}{1 + {3^x}}}} dx

Step 3: Simplify and evaluate the integral. $2I = \int\limits_{ - \pi /2}^{\pi /2} {{{\cos }^2}x} dx$ Using the identity $\cos^2 x = \frac{1+\cos(2x)}{2}$: $2I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{1+\cos(2x)}{2}} dx$ $2I = \frac{1}{2} \int\limits_{ - \pi /2}^{\pi /2} {(1+\cos(2x))} dx$ $2I = \frac{1}{2} \left[ x + \frac{1}{2}\sin(2x) \right]_{ - \pi /2}^{\pi /2}$ $2I = \frac{1}{2} \left[ \left( \frac{\pi}{2} + \frac{1}{2}\sin(\pi) \right) - \left( -\frac{\pi}{2} + \frac{1}{2}\sin(-\pi) \right) \right]$ $2I = \frac{1}{2} \left[ \left( \frac{\pi}{2} + 0 \right) - \left( -\frac{\pi}{2} + 0 \right) \right]$ $2I = \frac{1}{2} \left[ \frac{\pi}{2} + \frac{\pi}{2} \right] = \frac{1}{2} [\pi] = \frac{\pi}{2}$

Step 4: Solve for $I$. $2I = \frac{\pi}{2} \implies I = \frac{\pi}{4}$

Common Mistakes & Tips

• Algebraic Errors: Be very careful with algebraic manipulations, especially when dealing with exponents like $3^x$ and $3^{-x}$.
• Trigonometric Identities: Ensure correct application of trigonometric identities, such as the double angle formula for $\cos^2 x$.
• Symmetric Limits Property Application: This property is most effective when the integrand can be simplified by adding $f(x)$ and $f(a+b-x)$. If simplification doesn't occur, the property might not be the easiest approach.
• Integral of $\cos^2 x$: A common mistake is to miscalculate the integral of $\cos^2 x$ over symmetric intervals.

Summary

The problem involves a definite integral with symmetric limits. The key to solving this integral is to apply the property $\int_a^b f(x) \,dx = \int_a^b f(a+b-x) \,dx$. By applying this property and adding the original integral to the transformed integral, the term involving $3^x$ cancels out, simplifying the integrand to $\cos^2 x$. Evaluating the integral of $\cos^2 x$ over the given limits yields the final result. My derivation consistently shows the value of the integral to be $\frac{\pi}{4}$.

The final answer is \boxed{{\pi \over 4}}.