If $\int \frac{1}{\mathrm{x}} \sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}} \mathrm{dx}=\mathrm{g}(\mathrm{x})+\mathrm{c}, \mathrm{g}(1)=0$, then $\mathrm{g}\left(\frac{1}{2}\right)$ is equal to :

(A) $\log _{\mathrm{e}}\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)+\frac{\pi}{3}$

(B) $\log _{\mathrm{e}}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)+\frac{\pi}{3}$

(C) $\log _{\mathrm{e}}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)-\frac{\pi}{3}$

(D) $\frac{1}{2} \log _{\mathrm{e}}\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)-\frac{\pi}{6}$

Indefinite Integrals — JEE Main · 2022 · 26 Jun 2022 · Shift 2 Question with Solution

\int \frac{1}{x} \frac{1 - x}{1 + x} dx = g (x) + c, g (1) = 0

, then

g (\frac{1}{2})

is equal to :

Answer: (A)

Substitute

x = cos 2 θ

, so

d x = - 2 sin 2 θ d θ

\frac{1 - x}{1 + x} = \frac{1 - c o s 2 θ}{1 + c o s 2 θ} = \frac{2 s i n ^{2} θ}{2 c o s ^{2} θ} = tan θ

The integral becomes:

\int \frac{t a n θ}{c o s 2 θ} (- 2 sin 2 θ) d θ = - 2 \int \frac{2 s i n ^{2} θ}{c o s 2 θ} d θ = - 2 \int \frac{1 - c o s 2 θ}{c o s 2 θ} d θ

= - 2 \int (sec 2 θ - 1) d θ = - ln ∣ sec 2 θ + tan 2 θ ∣ + 2 θ + c

= ln ∣ sec 2 θ - tan 2 θ ∣ + 2 θ + c = ln \frac{1 - 1 - x ^{2}}{x} + cos^{- 1} x + c

Using

g (1) = 0

, we get

c = 0

, so:

g (x) = ln \frac{1 - 1 - x ^{2}}{x} + cos^{- 1} x

g (\frac{1}{2}) = ln \frac{1 - \frac{3}{2}}{\frac{1}{2}} + \frac{π}{3} = ln ∣2 - 3 ∣ + \frac{π}{3} = ln (\frac{3 - 1}{3 + 1}) + \frac{π}{3}

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option