Question:
The value of the integral \( \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5x}{1 + e^{5x}} \, dx \) is equal to:
The value of the integral \( \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5x}{1 + e^{5x}} \, dx \) is equal to:
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For any integral \( \int_{-a}^{a} \frac{f(x)}{1 + p^{g(x)}} \, dx \) where \( f(x) \) is even and \( g(x) \) is odd, the value is simply \( \int_{0}^{a} f(x) \, dx \).
Updated On: May 20, 2026
- \( \frac{1}{5} \)
- \( \frac{\sqrt{3}}{10} \)
- \( \frac{1}{15} \)
- \( \frac{1}{10} \)
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The Correct Option is B
Solution and Explanation
Concept:
The solution utilizes the property of definite integrals often referred to as King's Property. For an integral with symmetric limits \( [-a, a] \):
• \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \)
• This is particularly useful when the integrand contains an exponential term like \( e^{kx} \) in the denominator alongside an even function like \( \cos x \).
Step 1: Applying the integral property \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \).
Let \( I = \int_{-\pi/15}^{\pi/15} \frac{\cos 5x}{1 + e^{5x}} \, dx \quad \cdots (1) \)
Using the property, replace \( x \) with \( ( \frac{\pi}{15} - \frac{\pi}{15} - x ) = -x \): \[ I = \int_{-\pi/15}^{\pi/15} \frac{\cos 5(-x)}{1 + e^{5(-x)}} \, dx = \int_{-\pi/15}^{\pi/15} \frac{\cos 5x}{1 + e^{-5x}} \, dx \] Multiplying numerator and denominator by \( e^{5x} \): \[ I = \int_{-\pi/15}^{\pi/15} \frac{e^{5x} \cos 5x}{e^{5x} + 1} \, dx \quad \cdots (2) \]
Step 2: Adding equations (1) and (2).
\[ 2I = \int_{-\pi/15}^{\pi/15} \left( \frac{\cos 5x}{1 + e^{5x}} + \frac{e^{5x} \cos 5x}{1 + e^{5x}} \right) \, dx \] \[ 2I = \int_{-\pi/15}^{\pi/15} \frac{\cos 5x (1 + e^{5x})}{1 + e^{5x}} \, dx = \int_{-\pi/15}^{\pi/15} \cos 5x \, dx \]
Step 3: Evaluating the resulting integral.
Since \( \cos 5x \) is an even function: \[ 2I = 2 \int_{0}^{\pi/15} \cos 5x \, dx \quad \Rightarrow \quad I = \int_{0}^{\pi/15} \cos 5x \, dx \] \[ I = \left[ \frac{\sin 5x}{5} \right]_{0}^{\pi/15} = \frac{1}{5} \left( \sin \frac{5\pi}{15} - \sin 0 \right) \] \[ I = \frac{1}{5} \sin \frac{\pi}{3} = \frac{1}{5} \left( \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{10} \]
• \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \)
• This is particularly useful when the integrand contains an exponential term like \( e^{kx} \) in the denominator alongside an even function like \( \cos x \).
Step 1: Applying the integral property \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \).
Let \( I = \int_{-\pi/15}^{\pi/15} \frac{\cos 5x}{1 + e^{5x}} \, dx \quad \cdots (1) \)
Using the property, replace \( x \) with \( ( \frac{\pi}{15} - \frac{\pi}{15} - x ) = -x \): \[ I = \int_{-\pi/15}^{\pi/15} \frac{\cos 5(-x)}{1 + e^{5(-x)}} \, dx = \int_{-\pi/15}^{\pi/15} \frac{\cos 5x}{1 + e^{-5x}} \, dx \] Multiplying numerator and denominator by \( e^{5x} \): \[ I = \int_{-\pi/15}^{\pi/15} \frac{e^{5x} \cos 5x}{e^{5x} + 1} \, dx \quad \cdots (2) \]
Step 2: Adding equations (1) and (2).
\[ 2I = \int_{-\pi/15}^{\pi/15} \left( \frac{\cos 5x}{1 + e^{5x}} + \frac{e^{5x} \cos 5x}{1 + e^{5x}} \right) \, dx \] \[ 2I = \int_{-\pi/15}^{\pi/15} \frac{\cos 5x (1 + e^{5x})}{1 + e^{5x}} \, dx = \int_{-\pi/15}^{\pi/15} \cos 5x \, dx \]
Step 3: Evaluating the resulting integral.
Since \( \cos 5x \) is an even function: \[ 2I = 2 \int_{0}^{\pi/15} \cos 5x \, dx \quad \Rightarrow \quad I = \int_{0}^{\pi/15} \cos 5x \, dx \] \[ I = \left[ \frac{\sin 5x}{5} \right]_{0}^{\pi/15} = \frac{1}{5} \left( \sin \frac{5\pi}{15} - \sin 0 \right) \] \[ I = \frac{1}{5} \sin \frac{\pi}{3} = \frac{1}{5} \left( \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{10} \]
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