Q65 / 120 · VITEEE 2012 · Mathematics
The value of integral $\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x$ is
- A$\frac{\pi}{2}+1$
- Correct answerB$\frac{\pi}{2}-1$
- C$-1$
- D1
Solution
Let $\mathrm{I}=\int_{0}^{1} \sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}} \mathrm{dx}$
$=\int_{0}^{1} \frac{1-x}{\sqrt{1-x^{2}}} d x$
$=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} d x-\int_{0}^{1} \frac{x}{\sqrt{1+x^{2}}} d x$
$=\left[\sin ^{-1} x\right]_{0}^{1}-\int_{0}^{1} \frac{x}{\sqrt{1-x^{2}}} d x$
Put $t^{2}=1-x^{2} \Rightarrow 2 t d t=-2 x d x$
$\Rightarrow t \mathrm{dt}=-x \mathrm{~d} x$
$\therefore \mathrm{I}=\left(\sin ^{-1} 1-\sin ^{-1} 0\right)+\int_{1}^{0} \frac{\mathrm{t}}{\mathrm{t}} \mathrm{dt}$
$=\frac{\pi}{2}+[\mathrm{t}]_{1}^{0}=\frac{\pi}{2}-1$
Hence, the correct answer is option: 'B'.