If [ 1 x 1 - x 1 + x dx = g(x) + c ], g(1) = 0, then [g( 1 2 )] is equal to :

Question

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 :

Accepted Answer

Given, $$\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx = g(x) + c} $$, $g(1) = 0$

Let I = $\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx}$

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

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

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

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]

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

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

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

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)$

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

Solutions · 1