Anti-derivative of $\dfrac{\tan x-1}{\tan x+1}$ with respect to $x$ is:

(C) $\log\left|\sec\left(\frac{\pi}{4}-x\right)\right|+c$

(A) $\sec^{2}\left(\frac{\pi}{4}-x\right)+c$

(B) $-\sec^{2}\left(\frac{\pi}{4}-x\right)+c$

(D) $-\log\left|\sec\left(\frac{\pi}{4}-x\right)\right|+c$

CBSE Class 12 · 2023 · Indefinite IntegralsMultiple Choice

Anti-derivative of

\frac{tan x - 1}{tan x + 1}

with respect to

x

is:

Answer: (C)

Using

tan (\frac{π}{4} - x) = \frac{1 - tan x}{1 + tan x}

, the integrand is

\frac{tan x - 1}{tan x + 1} = - tan (\frac{π}{4} - x) .

Put

t = \frac{π}{4} - x

d t = - d x

\int - tan t (- d t) = \int tan t d t = lo g ∣ sec t ∣ + C .

Restoring

t

lo g sec (\frac{π}{4} - x) + C

Hence option (C).

CBSE Class 12 Board · 2023 · Single correct option