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

Step 1: Split the bracket.
\[ I=\int e^x\left(\frac{1}{n}\sec nx+\sec nx\tan nx\right)dx \]
Step 2: Recognise the special pattern.
The rule $\int e^x[f(x)+f'(x)]\,dx=e^xf(x)+c$ fits perfectly if we choose $f(x)=\frac{1}{n}\sec nx$.
Step 3: Check that $f'(x)$ matches.
\[ f'(x)=\frac{1}{n}\cdot n\sec nx\tan nx=\sec nx\tan nx \] which is exactly the second term, so the pattern holds.
Step 4: Write the integral.
\[ I=e^x\cdot\frac{1}{n}\sec nx=\frac{1}{n}e^x\sec nx \]
Step 5: Match the given form.
The problem writes the answer as $\frac{1}{n}(g(x)+k)$, so $g(x)=e^x\sec nx$ and $F(x)=\frac{1}{n}(e^x\sec nx+k)$.
Step 6: Use $F(0)=1$.
\[ 1=\frac{1}{n}(e^0\sec0+k)=\frac{1}{n}(1+k)\implies n=1+k\implies k=n-1 \] \[ \boxed{n-1} \]