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. $\sec x + \tan x + \dfrac{1}{2}$

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

Given equation is $\int {{e^{\sec x}}\left( {\sec x\tan xf(x) + (\sec x\tan x + {{\sec }^2}x)} \right)} dx = {e^{\sec x}}f(x) + C$
$${e^{\sec x}}\left( {\sec x\tan xf(x) + (\sec x\tan x + {{\sec }^2}x)} \right) = {e^{\sec x}} \cdot \sec x \cdot \tan x \cdot f(x) + {e^{\sec x}} \cdot f'(x)$$
$$f'(x) = {\sec ^2}x + \tan x \cdot \sec x$$
$$\Rightarrow \int {f'(x)} = \int {({{\sec }^2}x + \tan x \cdot \sec x} )dx$$
$$\Rightarrow f(x) = \tan x + \sec x + c$$
$\therefore$Option D is the correct answer.
$$\int {{{\sec }^2}x\;} dx = \tan x + c$$
$$\int {\tan x \cdot \sec x} \;dx = \sec x + c$$