Question:
The value of the integral
\[
\int_{-\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{\sec(2023x)}{1 + (2023)^{2023x}} \, dx
\]
is equal to:
The value of the integral
\[
\int_{-\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{\sec(2023x)}{1 + (2023)^{2023x}} \, dx
\]
is equal to:
Show Hint
The "King's Rule" effectively eliminates the non-symmetric part of the denominator (the exponential term) in symmetric intervals, reducing complex integrals to standard trigonometric forms.
Updated On: May 20, 2026
- \( \frac{1}{2023\sqrt{2}} + C \)
- \( \frac{\log(\sqrt{2} + 1)}{2023} + C \)
- \( \frac{\log 2}{4046} + C \)
- \( \frac{\sqrt{2}}{2023} + C \)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
For a definite integral with symmetric limits \( [-a, a] \), we use the property:
• \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \)
• In the form \( \int_{-a}^{a} \frac{f(x)}{1 + p^{g(x)}} \, dx \), if \( f(x) \) is an even function and \( g(x) \) is an odd function, the integral simplifies to \( \int_{0}^{a} f(x) \, dx \).
Step 1: Applying the integral property.
Let \( I = \int_{-\pi/8092}^{\pi/8092} \frac{\sec(2023x)}{1 + (2023)^{(2023x)}} \, dx \quad \cdots (1) \)
Replacing \( x \) with \( -x \): \[ I = \int_{-\pi/8092}^{\pi/8092} \frac{\sec(2023(-x))}{1 + (2023)^{(2023(-x))}} \, dx = \int_{-\pi/8092}^{\pi/8092} \frac{\sec(2023x)}{1 + (2023)^{-2023x}} \, dx \] Simplifying the denominator by multiplying top and bottom by \( (2023)^{2023x} \): \[ I = \int_{-\pi/8092}^{\pi/8092} \frac{(2023)^{2023x} \sec(2023x)}{1 + (2023)^{2023x}} \, dx \quad \cdots (2) \]
Step 2: Adding equations (1) and (2).
\[ 2I = \int_{-\pi/8092}^{\pi/8092} \frac{\sec(2023x) [1 + (2023)^{2023x}]}{1 + (2023)^{2023x}} \, dx \] \[ 2I = \int_{-\pi/8092}^{\pi/8092} \sec(2023x) \, dx \] Using the even property of the secant function (\( \sec(-x) = \sec x \)): \[ 2I = 2 \int_{0}^{\pi/8092} \sec(2023x) \, dx \quad \Rightarrow \quad I = \int_{0}^{\pi/8092} \sec(2023x) \, dx \]
Step 3: Integrating and applying limits.
Using the standard integral \( \int \sec(mx) \, dx = \frac{1}{m} \log|\sec mx + \tan mx| \): \[ I = \left[ \frac{1}{2023} \log|\sec(2023x) + \tan(2023x)| \right]_{0}^{\pi/8092} \] Substituting the upper limit \( (2023 \times \frac{\pi}{8092} = \frac{\pi}{4}) \) and lower limit \( 0 \): \[ I = \frac{1}{2023} \left[ \log(\sec \frac{\pi}{4} + \tan \frac{\pi}{4}) - \log(\sec 0 + \tan 0) \right] \] \[ I = \frac{1}{2023} [ \log(\sqrt{2} + 1) - \log(1 + 0) ] = \frac{\log(\sqrt{2} + 1)}{2023} \]
• \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \)
• In the form \( \int_{-a}^{a} \frac{f(x)}{1 + p^{g(x)}} \, dx \), if \( f(x) \) is an even function and \( g(x) \) is an odd function, the integral simplifies to \( \int_{0}^{a} f(x) \, dx \).
Step 1: Applying the integral property.
Let \( I = \int_{-\pi/8092}^{\pi/8092} \frac{\sec(2023x)}{1 + (2023)^{(2023x)}} \, dx \quad \cdots (1) \)
Replacing \( x \) with \( -x \): \[ I = \int_{-\pi/8092}^{\pi/8092} \frac{\sec(2023(-x))}{1 + (2023)^{(2023(-x))}} \, dx = \int_{-\pi/8092}^{\pi/8092} \frac{\sec(2023x)}{1 + (2023)^{-2023x}} \, dx \] Simplifying the denominator by multiplying top and bottom by \( (2023)^{2023x} \): \[ I = \int_{-\pi/8092}^{\pi/8092} \frac{(2023)^{2023x} \sec(2023x)}{1 + (2023)^{2023x}} \, dx \quad \cdots (2) \]
Step 2: Adding equations (1) and (2).
\[ 2I = \int_{-\pi/8092}^{\pi/8092} \frac{\sec(2023x) [1 + (2023)^{2023x}]}{1 + (2023)^{2023x}} \, dx \] \[ 2I = \int_{-\pi/8092}^{\pi/8092} \sec(2023x) \, dx \] Using the even property of the secant function (\( \sec(-x) = \sec x \)): \[ 2I = 2 \int_{0}^{\pi/8092} \sec(2023x) \, dx \quad \Rightarrow \quad I = \int_{0}^{\pi/8092} \sec(2023x) \, dx \]
Step 3: Integrating and applying limits.
Using the standard integral \( \int \sec(mx) \, dx = \frac{1}{m} \log|\sec mx + \tan mx| \): \[ I = \left[ \frac{1}{2023} \log|\sec(2023x) + \tan(2023x)| \right]_{0}^{\pi/8092} \] Substituting the upper limit \( (2023 \times \frac{\pi}{8092} = \frac{\pi}{4}) \) and lower limit \( 0 \): \[ I = \frac{1}{2023} \left[ \log(\sec \frac{\pi}{4} + \tan \frac{\pi}{4}) - \log(\sec 0 + \tan 0) \right] \] \[ I = \frac{1}{2023} [ \log(\sqrt{2} + 1) - \log(1 + 0) ] = \frac{\log(\sqrt{2} + 1)}{2023} \]
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