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

Question

Accepted Answer

Answer: A. {\log _e}\left( {{{\sqrt 3 - 1} \over {\sqrt 3 + 1}}} ight) + {\pi \over 3} Given, \int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx = 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} ight)\left( {1 + x} ight)} \over {\left( {1 + x} ight)\left( {1 + x} ight)}}} } 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}ight|-\sin ^{-1}(x)+C_1 =-\ln \left|\frac{1}{\mathrm{x}}+\sqrt{\frac{1}{\mathrm{x}^2}-1}ight|-\sin ^{-1}(\mathrm{x})+\mathrm{C}_1 [Putting values of t] =-\ln \left|\frac{1}{\mathrm{x}}+\sqrt{\frac{1}{\mathrm{x}^2}-1}ight|-\left(\frac{\pi}{2}-\cos ^{-1} \mathrm{x}ight)+\mathrm{C}_1 =-\ln \left|\frac{1}{\mathrm{x}}+\sqrt{\frac{1}{\mathrm{x}^2}-1}ight|+\cos ^{-1}(\mathrm{x})-\frac{\pi}{2}+\mathrm{C}_1 	herefore g(x)=\cos ^{-1}(x)-\ell n\left|\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}ight| So, g(1)=\cos ^{-1}(1)-\ell n|1|=0 g\left(\frac{1}{2}ight)=\cos ^{-1}\left(\frac{1}{2}ight)-\ell n|2+\sqrt{3}| = {\pi \over 3} + \ln \left( {{1 \over {2 + \sqrt 3 }}} ight) =\frac{\pi}{3}+\ln \left(\frac{\sqrt{3}-1}{\sqrt{3}+1}ight)

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

Answer

Explanation

More JEE Math questions