JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 13)

$$\int {{e^{\sec x}}}$$ $$(\sec x\tan xf(x) + \sec x\tan x + se{x^2}x)dx$$
= esecxf(x) + C then a possible choice of f(x) is :-
x sec x + tan x + 1/2
sec x + xtan x - 1/2
sec x - tan x - 1/2
sec x + tan x + 1/2

Explanation

Given

$$\int {{e^{\sec x}}\left( {\sec x\tan x\,f(x) + (sec\,x\,tan\,x\, + se{c^2}x)} \right)} $$

$$ = {e^{\sec x}}f(x) + C$$

Differentiating both sides with respect to x,

$${e^{\sec x}}.\sec x\tan x\,f\left( x \right)$$ + $${e^{\sec x}}\left( {\sec x\tan x + {{\sec }^2}x} \right)$$

= $${e^{\sec x}}$$. f'(x) + f(x)$${e^{\sec x}}$$.$$\sec x\tan x$$

$$ \Rightarrow $$ $${e^{\sec x}}\left( {\sec x\tan x + {{\sec }^2}x} \right)$$ = $${e^{\sec x}}$$. f'(x)

$$ \Rightarrow $$ f'(x) = $${\sec x\tan x + {{\sec }^2}x}$$

$$ \therefore $$ $$f(x) = \int {\left( {\left( {\sec x\tan x} \right) + {{\sec }^2}x} \right)dx} $$

$$ \therefore $$ $$f(x) = \sec x + \tan x + C$$

From question we can not find the value of C. So we have to choose any random value of C where ($$\sec x + \tan x$$) present.

Only in option D, ($$\sec x + \tan x$$) term present. So only possible option is D.

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