Question

The value of integral $\int\limits_0^1 \, \sqrt{\frac{1-x}{1+x}}dx$ is

• A. $\frac{\pi}{2} + 1$
• B. $\frac{\pi}{2} - 1$
• C. $-1$
• D. $1$

Answer 1

The Correct Option is B

Let $I = \int\limits^{1}_{0} \sqrt{\frac{1-x}{1+x}} dx$
$ = \int\limits^{1}_{0} \frac{1-x}{\sqrt{1-x^{2}}}dx$
$ = \int\limits^{1}_{0} \frac{1}{\sqrt{1-x^{2}}} dx - \int\limits^{1}_{0} \frac{x}{\sqrt{1+x^{2}}}dx $
$= \left[\sin^{-1}x\right]^{1}_{0} - \int\limits^{1}_{0} \frac{x}{\sqrt{1-x^{2}}} dx$
Put $ t^{2} = 1 -x^{2} $
$ \Rightarrow \, 2t \, dt = - 2x \, dx $
$\Rightarrow \, t \, dt = -x \, dx $
$\therefore I = \left(\sin^{-1} 1 - \sin^{-1}0\right)+\int\limits_{1}^{0} \frac{t}{t} dt$
$ = \frac{\pi}{2} + \left[t\right]^{0}_{1} = \frac{\pi}{2} - 1 $
Let $I=\int\limits_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x$
Put $x=\cos 2 \,\theta$
$ \Rightarrow d x=-2 \sin 2 \,\theta \,d \,\theta$
$\therefore I=-\int\limits_{\pi / 4}^{0} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}} \cdot 2 \sin \theta \,d\, \theta$
$=-2 \int\limits_{\pi / 4}^{0} \frac{\sin \theta}{\cos \theta} \cdot 2 \sin \theta \cos \theta \,d \,\theta$
$=-2 \int\limits_{\pi / 4}^{0}(1-\cos 2 \theta) \,d \,\theta$
$=-2\left[\theta-\frac{\sin 2 \theta}{2}\right]_{\pi / 4}^{0}$
$=-2\left[0-\left(\frac{\pi}{4}-\frac{1}{2}\right)\right]$
$=\frac{\pi}{2}-1$