Step 1: Understanding the Concept:
The evaluation of definite integrals involving symmetric limits \([-a, a]\) and a denominator consisting of an exponential function alongside a trigonometric function is a classic problem in competitive mathematics.
The primary technique used here is the "King’s Property" or the reflection property of definite integrals.
This property allows us to transform the integrand in a way that the exponential term, which usually complicates direct integration, can be eliminated through addition.
The concept relies on the fact that for any continuous function \(f(x)\), the area under the curve from \(a\) to \(b\) is identical to the area under the reflected function \(f(a+b-x)\) over the same interval.
In this specific problem, we observe a symmetric interval and an even function (\(\cos 5x\)) in the numerator, paired with a non-symmetric exponential factor in the denominator.
Key Formula or Approach:
The fundamental property utilized is:
\[ \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx \]
For symmetric limits where \(b = -a\), the sum \(a + b = 0\), so the substitution becomes \(x \to -x\).
Additionally, we leverage the algebraic identity for even functions: \(f(-x) = f(x)\) for \(\cos(kx)\).
The strategy is to define the integral as \(I\), apply the property to get a second form of \(I\), and then add the two equations together.
Step 2: Detailed Explanation:
Let the given integral be denoted by \(I\):
\[ I = \int_{-\pi/15}^{\pi/15} \frac{\cos 5x}{1+e^{5x}} dx \quad \dots \text{ (Equation 1)} \]
Applying the integral property \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx\), we replace every instance of \(x\) with \((-\pi/15 + \pi/15 - x) = -x\).
\[ I = \int_{-\pi/15}^{\pi/15} \frac{\cos(5(-x))}{1+e^{5(-x)}} dx \]
Since the cosine function is even, \(\cos(-5x) = \cos(5x)\).
The denominator contains \(e^{-5x}\), which can be written as \(\frac{1}{e^{5x}}\).
Substituting these into the integral:
\[ I = \int_{-\pi/15}^{\pi/15} \frac{\cos 5x}{1 + \frac{1}{e^{5x}}} dx \]
To simplify this, we multiply the numerator and the denominator by \(e^{5x}\):
\[ I = \int_{-\pi/15}^{\pi/15} \frac{e^{5x} \cos 5x}{e^{5x} + 1} dx \quad \dots \text{ (Equation 2)} \]
Now, we add Equation 1 and Equation 2:
\[ I + I = \int_{-\pi/15}^{\pi/15} \left[ \frac{\cos 5x}{1+e^{5x}} + \frac{e^{5x} \cos 5x}{e^{5x} + 1} \right] dx \]
\[ 2I = \int_{-\pi/15}^{\pi/15} \frac{\cos 5x (1 + e^{5x})}{1 + e^{5x}} dx \]
Notice that the term \((1 + e^{5x})\) appears in both the numerator and denominator, so they cancel out completely:
\[ 2I = \int_{-\pi/15}^{\pi/15} \cos 5x dx \]
Because \(\cos 5x\) is an even function, the integral from \(-a\) to \(a\) is twice the integral from \(0\) to \(a\):
\[ 2I = 2 \int_{0}^{\pi/15} \cos 5x dx \implies I = \int_{0}^{\pi/15} \cos 5x dx \]
Integrating \(\cos 5x\) with respect to \(x\) gives \(\frac{\sin 5x}{5}\).
Applying the limits:
\[ I = \left[ \frac{\sin 5x}{5} \right]_{0}^{\pi/15} \]
\[ I = \frac{1}{5} \left[ \sin(5 \cdot \frac{\pi}{15}) - \sin(0) \right] \]
\[ I = \frac{1}{5} \left[ \sin(\frac{\pi}{3}) - 0 \right] \]
Since \(\sin(\pi/3) = \sqrt{3}/2\):
\[ I = \frac{1}{5} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{10} \]
Step 3: Final Answer:
After rigorous algebraic and trigonometric simplification using the King's Rule, the integral value evaluates to \(\frac{\sqrt{3}}{10}\).
This confirms the validity of option (B).