Question

If $\int {{e^{\sec x}}\left( {\sec x\tan xf(x) + (\sec x\tan x + {{\sec }^2}x)} \right)} dx = {e^{\sec x}}f(x) + C,$ find a possible choice of f(x)
A. $\sec x - \tan x - \dfrac{1}{2}$
B. $x\sec x + \tan x + \dfrac{1}{2}$
C. $\sec x + x\tan x - \dfrac{1}{2}$
D. $\sec x + \tan x + \dfrac{1}{2}$

Answer 1

Hint: We need to know the formulae of integration of basic trigonometric functions to solve the given problem.

Complete step-by-step answer:
Given equation is $\int {{e^{\sec x}}\left( {\sec x\tan xf(x) + (\sec x\tan x + {{\sec }^2}x)} \right)} dx = {e^{\sec x}}f(x) + C$
Differentiating the above equation both sides with respect to x,
$${e^{\sec x}}\left( {\sec x\tan xf(x) + (\sec x\tan x + {{\sec }^2}x)} \right) = {e^{\sec x}} \cdot \sec x \cdot \tan x \cdot f(x) + {e^{\sec x}} \cdot f'(x)$$
Cancelling the common terms on both sides of the above equation, we get
$$f'(x) = {\sec ^2}x + \tan x \cdot \sec x$$
We need to find f(x), so integrating the above equation with respect to x,
$$ \Rightarrow \int {f'(x)} = \int {({{\sec }^2}x + \tan x \cdot \sec x} )dx$$
$$ \Rightarrow f(x) = \tan x + \sec x + c$$
$\therefore $Option D is the correct answer.

Note: We need the value of f(x) from the given equation, for simplifying, we differentiate the given equation to get rid of extra terms and then again integrate to get the desired result. We used these basic integration formulae
$$\int {{{\sec }^2}x\;} dx = \tan x + c$$
$$\int {\tan x \cdot \sec x} \;dx = \sec x + c$$