If $\int e^{\sec x}\left(\sec x+\tan x f(x)+\left(\sec x \tan x+\sec ^{2} x\right)\right) d x=e^{\sec x} f(x)+C$, then a possible choice of $f(x)$ is

(D) $\sec x+\tan x+\frac{1}{2}$

(A) $\sec x-\tan x-\frac{1}{2}$

(B) $x \sec x+\tan x+\frac{1}{2}$

(C) $\sec x+x \tan x-\frac{1}{2}$

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Indefinite Integrals

\int e^{s e c x} (sec x + tan x f (x) + (sec x tan x + sec^{2} x)) d x = e^{s e c x} f (x) + C

, then a possible choice of

f (x)

Answer: (D)

The integrand is built so that the whole integral equals

e^{s e c x} f (x) + C

, so differentiating that right side must reproduce it. By the product rule,

\frac{d}{d x} (e^{s e c x} f (x)) = e^{s e c x} sec x tan x f (x) + e^{s e c x} f^{'} (x)

, and matching this against the integrand forces

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

. Integrating,

f (x) = tan x + sec x + C

, so a valid choice is

f (x) = sec x + tan x + \frac{1}{2}

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option