Question

If \( \int e^{x}\left(\frac{1}{n}+\tan nx\right)\sec nx \, dx = \frac{1}{n}(g(x)+k) = F(x) \) and \( F(0)=1 \), then \( k = \)

• A. \( n \)
• B. \( n+1 \)
• C. \( n-1 \)
• D. \( 1 \)

Hint:

Always look out for the \( \int e^x [f(x) + f'(x)] dx \) pattern whenever an exponent \( e^x \) is multiplied by trigonometric functions. It allows you to write down the final integration result instantly without using integration by parts.

Answer 1

The Correct Option is C

Concept: Let us rewrite the integral expression by distributing the terms inside the parentheses: \[ I = \int e^x \left( \frac{1}{n}\sec nx + \sec nx \tan nx \right) dx \] This perfectly matches the classic integral identity \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + c \). Let us choose: \[ f(x) = \frac{1}{n}\sec nx \implies f'(x) = \frac{1}{n}(\sec nx \tan nx \cdot n) = \sec nx \tan nx \]

Step 1: Evaluating the integral result.
Since it fits the classic identity perfectly, the result of the integration is: \[ I = e^x \left(\frac{1}{n}\sec nx\right) = \frac{1}{n} e^x \sec nx \]

Step 2: Matching with the given format to identify \( g(x) \).
The problem statement writes the answer as \( \frac{1}{n}(g(x) + k) \). Comparing the two forms: \[ g(x) = e^x \sec nx \] So the full function is \( F(x) = \frac{1}{n}(e^x \sec nx + k) \).

Step 3: Using the initial condition to solve for \( k \).
We are given \( F(0) = 1 \). Let us substitute \( x = 0 \) into our function: \[ 1 = \frac{1}{n}(e^0 \sec(0) + k) \implies 1 = \frac{1}{n}(1 + k) \] Cross-multiplying by \( n \): \[ n = 1 + k \implies k = n - 1 \] Let us re-verify options indexing. The analytical value evaluates exactly to \( n-1 \), which matches option (C).