QJEEM22Ma75d8: JEE Main 2022 Mathematics - Class 12 JEEM22Ma75d8 Solution: If ∫ 1 x 1 − x 1 + x d x = g ( x ) + c , g ( 1 ) = 0 , then g ( …

Question

Verified Solution • 2025–26 Session moderate

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

Accepted Answer

Given,

\int \frac{1}{x} \sqrt{\frac{1 - x}{1 + x}} d x = g (x) + c

,

g (1) = 0

Let I =

\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx}

=

\int {{1 \over x}\sqrt {{{\left( {1 - x} \right)\left( {1 + x} \right)} \over {\left( {1 + x} \right)\left( {1 + x} \right)}}} } dx

=\int \frac{1-x}{x \sqrt{1-x^2}} d x

=\int \frac{1}{x \sqrt{1-x^2}} d x-\int \frac{1}{\sqrt{1-x^2}} d x

Put

x=\frac{1}{t}

d x=-\frac{1}{t^2}

=\int \frac{\frac{-1}{t^2}}{\frac{1}{t} \sqrt{1-\frac{1}{t^2}}} d t-\sin ^{-1}(x)+C_1

=\int \frac{-d t}{\sqrt{t^2-1}}-\sin ^{-1}(x)+C_1

=-\ln \left|t+\sqrt{t^2-1}\right|-\sin ^{-1}(x)+C_1

=-\ln \left|\frac{1}{\mathrm{x}}+\sqrt{\frac{1}{\mathrm{x}^2}-1}\right|-\sin ^{-1}(\mathrm{x})+\mathrm{C}_1

[Putting values of t]

=-\ln \left|\frac{1}{\mathrm{x}}+\sqrt{\frac{1}{\mathrm{x}^2}-1}\right|-\left(\frac{\pi}{2}-\cos ^{-1} \mathrm{x}\right)+\mathrm{C}_1

=-\ln \left|\frac{1}{\mathrm{x}}+\sqrt{\frac{1}{\mathrm{x}^2}-1}\right|+\cos ^{-1}(\mathrm{x})-\frac{\pi}{2}+\mathrm{C}_1

∴

g(x)=\cos ^{-1}(x)-\ell n\left|\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right|

So,

g(1)=\cos ^{-1}(1)-\ell n|1|=0

g\left(\frac{1}{2}\right)=\cos ^{-1}\left(\frac{1}{2}\right)-\ell n|2+\sqrt{3}|

= {\pi \over 3} + \ln \left( {{1 \over {2 + \sqrt 3 }}} \right)

=\frac{\pi}{3}+\ln \left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)

Step-by-Step Verified Solution