Question

If \( \int e^{\sin x}(1 + \sec x \tan x)\, dx = e^{\sin x}f(x) + c \), then in \( 0 \leq x \leq 2\pi \), the number of solutions of \( f(x) = 1 \) is

• A. 4
• B. 0
• C. 2
• D. 3

Hint:

Always check if the derivative of a known product simplifies to the given integrand.

Answer 1

The Correct Option is C

Given: \( \int e^{\sin x}(1 + \sec x \tan x)dx = e^{\sin x}f(x) + c \).
By inspection and knowledge of standard integration rules, if we let: \[ f(x) = \tan x \Rightarrow f'(x) = \sec^2 x \] Try substitution \( u = \sin x + \ln |\sec x| \), then: \[ \frac{d}{dx}(e^{\sin x} \sec x) = e^{\sin x}( \cos x \sec x + \sec x \tan x ) = e^{\sin x}(1 + \sec x \tan x) \] Thus, \( f(x) = \sec x \). Solve \( \sec x = 1 \Rightarrow x = 0, 2\pi \), both lie in \( [0, 2\pi] \).
So, total number of solutions is 2.