If $f^{\prime}(x)=\tan ^{-1}(\sec x+\tan x),-\frac{\pi}{2}<x<\frac{\pi}{2}$, and $f(0)=0$, then $f(1)$ is equal to:

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Indefinite Integrals

f^{'} (x) = tan^{- 1} (sec x + tan x), - \frac{π}{2} < x < \frac{π}{2}

, and

f (0) = 0

, then

f (1)

is equal to:

Answer: (B)

Simplify the argument first:

sec x + tan x = \frac{1 + s i n x}{c o s x} = tan (\frac{π}{4} + \frac{x}{2})

. For

- \frac{π}{2} < x < \frac{π}{2}

the angle

\frac{π}{4} + \frac{x}{2}

lies in

(0, \frac{π}{2})

, so

f^{'} (x) = tan^{- 1} (tan (\frac{π}{4} + \frac{x}{2})) = \frac{π}{4} + \frac{x}{2}

. Integrating and using

f (0) = 0

gives

f (x) = \frac{π x}{4} + \frac{x ^{2}}{4}

, so

f (1) = \frac{π + 1}{4}

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option

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