JEE PYQ: Indefinite Integration Question 16
Question 16 - 2020 (09 Jan Shift 1)
If $f’(x) = \tan^{-1}(\sec x + \tan x)$, $-\frac{\pi}{2} < x < \frac{\pi}{2}$, and $f(0) = 0$, then $f(1)$ is equal to:
(1) $\frac{\pi+1}{4}$
(2) $\frac{1}{4}$
(3) $\frac{\pi-1}{4}$
(4) $\frac{\pi+2}{4}$
Show Answer
Answer: (1)
Solution
$f’(x) = \tan^{-1}\left(\frac{1+\sin x}{\cos x}\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right) = \frac{\pi}{4} + \frac{x}{2}$. Integrating: $f(x) = \frac{\pi x}{4} + \frac{x^2}{4} + C$. $f(0) = 0 \Rightarrow C = 0$. $f(1) = \frac{\pi+1}{4}$.
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