IIT-JEE 2004 ScreeningQuestion 1
The value of the integral $\displaystyle\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}\,dx$ is
Definite Integration
Solution
Concept. $x=\cos2\theta$ turns $\frac{1-x}{1+x}$ into $\tan^2\theta$. Formulas used. (1) $\frac{1-\cos2\theta}{1+\cos2\theta}=\frac{2\sin^2\theta}{2\cos^2\theta}=\tan^2\theta$(2) $dx=-2\sin2\theta\,d\theta$; $x:0\to1$ gives $\theta:\frac\pi4\to0$Step 1 - Substitute. By (1), (2), with $\tan\theta\ge0$ on $[0,\frac\pi4]$, $$I=\int_0^{\pi/4}\tan\theta\cdot2\sin2\theta\,d\theta=\int_0^{\pi/4}4\sin^2\theta\,d\theta.$$ Step 2 - Integrate. $4\sin^2\theta=2-2\cos2\theta$: $$I=\Big[2\theta-\sin2\theta\Big]_0^{\pi/4}=\frac\pi2-1.$$ Answer: (B)