JEE Main 2022 (Online) 26th June Evening ShiftQuestion 1
If $\int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}dx=g(x)+c$, $g(1)=0$, then $g\left(\frac{1}{2}\right)$ is equal to :
Indefinite Integration
Solution
Concept. $x=\cos\theta$ makes $\sqrt{\frac{1-x}{1+x}}=\tan\frac\theta2$, and the integrand becomes $-(\sec\theta-1)$. Formulas used. (1) $\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}=\tan\dfrac\theta2$, $\dfrac{dx}x=-\tan\theta\,d\theta$(2) $\tan\frac\theta2\tan\theta=\sec\theta-1$Step 1 - Substitute. By (1), (2), $$\int\frac1x\sqrt{\frac{1-x}{1+x}}dx=-\int(\sec\theta-1)\,d\theta=\theta-\ln|\sec\theta+\tan\theta|+c.$$ Step 2 - Constant. $x=1$ means $\theta=0$: $g(1)=0-\ln1+c=0$, so $c=0$. Step 3 - $x=\frac12$. $\theta=\frac\pi3$: $g=\frac\pi3-\ln(2+\sqrt3)$. Step 4 - Match the option. $\frac{\sqrt3-1}{\sqrt3+1}=\frac{(\sqrt3-1)^2}2=2-\sqrt3=\frac1{2+\sqrt3}$, so $$g\Big(\frac12\Big)=\log_e\frac{\sqrt3-1}{\sqrt3+1}+\frac\pi3.$$ Answer: (A)