Question

Evaluate the integral: \[ I = \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5x}{1 + e^{5x}} \, dx \] 

• A. \( \frac{1}{5} \)
• B. \( \frac{\sqrt{3}}{10} \)
• C. \( \frac{1}{15} \)
• D. \( \frac{1}{10} \)

Hint:

For definite integrals with limits \([-a, a]\), use the property \( f(x) + f(-x) \) to simplify the expression. Also, keep in mind the even nature of the cosine function.

Answer 1

The Correct Option is B

Step 1: Using the Property \( f(x) + f(-x) \) Define: \[ I = \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5x}{1 + e^{5x}} \, dx \] Using the property: \[ \int_{-a}^{a} f(x) \, dx = \int_{-a}^{a} f(-x) \, dx \] Substituting \( x \to -x \), we get: \[ I = \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{\cos(-5x)}{1 + e^{-5x}} \, dx \] Since \( \cos(-5x) = \cos 5x \), we rewrite: \[ I = \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5x}{1 + e^{-5x}} \, dx \] Using the property \( e^{-5x} = \frac{1}{e^{5x}} \), we obtain: \[ \frac{\cos 5x}{1 + e^{-5x}} = \frac{\cos 5x}{1 + \frac{1}{e^{5x}}} \] Multiplying numerator and denominator by \( e^{5x} \), we get: \[ = \frac{e^{5x} \cos 5x}{e^{5x} + 1} \] Now, add the two equations for \( I \): \[ I + I = \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5x}{1 + e^{5x}} \, dx + \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{e^{5x} \cos 5x}{e^{5x} + 1} \, dx \] Since the integrals are equal in magnitude, the sum simplifies and gives us: \[ 2I = \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \cos 5x \, dx \] The integral of \( \cos 5x \) over symmetric limits \([-a, a]\) is zero, so we get: \[ I = \frac{\sqrt{3}}{10} \]